OP Malhotra Class 11 Maths Solutions Chapter 22 Probability Exercise 22 (D)

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(d)

Question 1. Four digit numbers are formed by using the digits 1, 2, 3, 4 and 5 without repeating the digit. Find the probability that a number, chose at random, is an odd number.
Answer: First, we find the total number of 4-digit numbers that can be made using the digits 1, 2, 3, 4, and 5 without repeating any digit. For the first digit, there are 5 choices. For the second, 4 choices remain. For the third, 3 choices, and for the fourth, 2 choices. So, the total number of possible 4-digit numbers is \( 5 \times 4 \times 3 \times 2 = 120 \). These are all the possible outcomes.
Next, we need to find the number of these 4-digit numbers that are odd. An odd number must end with an odd digit. The given odd digits are 1, 3, and 5.
So, for the unit's place (U), there are 3 choices (1, 3, or 5).
For the thousands place (Th), there are 4 remaining choices (since one digit is used for the unit's place).
For the hundreds place (H), there are 3 remaining choices.
For the tens place (T), there are 2 remaining choices.
Thus, the total number of 4-digit odd numbers is \( 4 \times 3 \times 2 \times 3 = 72 \). This is the number of favorable outcomes.
The probability is found by dividing the number of favorable outcomes by the total number of outcomes.
Required probability \( = \frac{\text{Total number of favourable cases}}{\text{Total number of cases}} = \frac{72}{120} = \frac{3}{5} \).
In simple words: We first count all possible 4-digit numbers you can make. Then, we count how many of those numbers are odd. Finally, we divide the count of odd numbers by the total count to get the probability. Every digit can only be used once.

🎯 Exam Tip: When forming numbers with restrictions, always start by fulfilling the restriction (like the odd digit for the unit's place) before filling the other places.

Question 2. What is the probability of getting 3 white balls in a draw of 3 balls from a box containing 6 white and 4 red balls ?
Answer: First, we find the total number of balls in the box. There are 6 white balls and 4 red balls, making a total of \( 6 + 4 = 10 \) balls.
We are drawing 3 balls from these 10 balls. The total number of ways to draw 3 balls from 10 is given by the combination formula \( ^{n}C_{r} \), which is \( ^{10}C_{3} \).
\( ^{10}C_{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120 \). So, there are 120 possible ways to draw 3 balls.
Next, we want to find the number of ways to get exactly 3 white balls. There are 6 white balls in total, and we want to choose 3 of them. This is given by \( ^{6}C_{3} \).
\( ^{6}C_{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \). So, there are 20 ways to draw 3 white balls.
The required probability is the number of favorable outcomes (drawing 3 white balls) divided by the total number of possible outcomes (drawing any 3 balls).
Required probability \( = \frac{^{6}C_{3}}{^{10}C_{3}} = \frac{20}{120} = \frac{1}{6} \).
In simple words: You have 10 balls, 6 white and 4 red. You pick 3 balls. We want to know the chance that all 3 balls you pick are white. First, count all the ways to pick any 3 balls. Then, count all the ways to pick only 3 white balls. Divide the second number by the first number to get the probability.

🎯 Exam Tip: Remember to use combinations \( (^{n}C_{r}) \) when the order of selection doesn't matter, as is the case when drawing balls from a box.

Question 3. A bag contains 6 white, 7 red and 5 black balls, three balls are drawn at random. Find the probability that they will be white.
Answer: First, find the total number of balls in the bag: \( 6 \text{ (white)} + 7 \text{ (red)} + 5 \text{ (black)} = 18 \) balls.
We are drawing 3 balls at random from these 18 balls. The total number of possible ways to do this is \( ^{18}C_{3} \).
\( ^{18}C_{3} = \frac{18!}{3!(18-3)!} = \frac{18!}{3!15!} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 3 \times 17 \times 16 = 816 \).
Next, we want to find the number of ways that all three drawn balls are white. There are 6 white balls in the bag, and we need to choose 3 of them. This is \( ^{6}C_{3} \).
\( ^{6}C_{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \).
The probability of drawing 3 white balls is the number of favorable outcomes divided by the total number of outcomes.
Required probability \( = \frac{^{6}C_{3}}{^{18}C_{3}} = \frac{20}{816} = \frac{5}{204} \). It's helpful to simplify fractions as much as possible.
In simple words: We have a bag with different colored balls. We pick 3 balls without looking. We want to know the chance that all 3 balls are white. First, count how many ways you can pick any 3 balls from the bag. Then, count how many ways you can pick 3 white balls. Divide these two numbers to get the answer.

🎯 Exam Tip: Always make sure to calculate both the total number of possible outcomes (exhaustive cases) and the number of desired outcomes (favorable cases) correctly before finding the ratio for probability.

Question 4. A bag contains 2 white marbles, 4 blue marbles, and 6 red marbles. A marble is drawn at random from the bag. What is the probability' that
(i) it is white ?
(ii) it is blue ?
(iii) it is red ?
(iv) it is not white ?
(v) it is not blue ?
(vi) it is black ?
Answer: First, let's find the total number of marbles in the bag: \( 2 \text{ (white)} + 4 \text{ (blue)} + 6 \text{ (red)} = 12 \) marbles. This is the total number of exhaustive cases when drawing one marble.
(i) Probability of drawing a white marble:
There are 2 white marbles.
\( P(\text{white}) = \frac{\text{Number of white marbles}}{\text{Total number of marbles}} = \frac{2}{12} = \frac{1}{6} \)
(ii) Probability of drawing a blue marble:
There are 4 blue marbles.
\( P(\text{blue}) = \frac{\text{Number of blue marbles}}{\text{Total number of marbles}} = \frac{4}{12} = \frac{1}{3} \)
(iii) Probability of drawing a red marble:
There are 6 red marbles.
\( P(\text{red}) = \frac{\text{Number of red marbles}}{\text{Total number of marbles}} = \frac{6}{12} = \frac{1}{2} \)
(iv) Probability of drawing a marble that is not white:
This means drawing either a red or a blue marble.
Number of non-white marbles = Number of red marbles + Number of blue marbles \( = 6 + 4 = 10 \).
\( P(\text{not white}) = \frac{10}{12} = \frac{5}{6} \). Alternatively, \( P(\text{not white}) = 1 - P(\text{white}) = 1 - \frac{1}{6} = \frac{5}{6} \).
(v) Probability of drawing a marble that is not blue:
This means drawing either a white or a red marble.
Number of non-blue marbles = Number of white marbles + Number of red marbles \( = 2 + 6 = 8 \).
\( P(\text{not blue}) = \frac{8}{12} = \frac{2}{3} \).
(vi) Probability of drawing a black marble:
There are no black marbles in the bag. So, the number of favorable cases is 0.
\( P(\text{black}) = \frac{0}{12} = 0 \). It's impossible to draw a black marble if none are present.
In simple words: First, count all the marbles. Then, for each color, divide the number of marbles of that color by the total number of marbles to find its chance. If you want the chance of "not" a color, count all the other colors and divide by the total, or subtract the chance of that color from 1. If a color is not in the bag, its chance is zero.

🎯 Exam Tip: When finding the probability of "not A," it's often simpler to calculate \( 1 - P(A) \) than to count all other possibilities, especially for mutually exclusive events.

Question 5. In Q. 4, three marbles are drawn from the bag. What is the probability that
(i) they are all blue ?
(ii) they are all red ?
(iii) they are all white ?
(iv) none of them is red ?
(v) not all of them are red ?
(vi) none is black ?
Answer: From Question 4, the bag contains:
2 white marbles
4 blue marbles
6 red marbles
Total number of marbles = 12.
We are drawing 3 marbles at random.
Total number of exhaustive cases (ways to draw 3 marbles from 12) is \( ^{12}C_{3} \).
\( ^{12}C_{3} = \frac{12!}{3!9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220 \).
(i) Probability that all three marbles are blue:
There are 4 blue marbles. Number of ways to choose 3 blue marbles is \( ^{4}C_{3} \).
\( ^{4}C_{3} = \frac{4!}{3!1!} = 4 \).
Required probability \( = \frac{^{4}C_{3}}{^{12}C_{3}} = \frac{4}{220} = \frac{1}{55} \).
(ii) Probability that all three marbles are red:
There are 6 red marbles. Number of ways to choose 3 red marbles is \( ^{6}C_{3} \).
\( ^{6}C_{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \).
Required probability \( = \frac{^{6}C_{3}}{^{12}C_{3}} = \frac{20}{220} = \frac{1}{11} \).
(iii) Probability that all three marbles are white:
There are only 2 white marbles. It's impossible to draw 3 white marbles.
Number of ways to choose 3 white marbles is \( ^{2}C_{3} = 0 \).
Required probability \( = \frac{0}{220} = 0 \).
(iv) Probability that none of them are red:
This means all 3 marbles drawn must be from the non-red marbles.
Number of non-red marbles = Total marbles - Number of red marbles \( = 12 - 6 = 6 \) (these are white and blue marbles).
Number of ways to choose 3 marbles from these 6 non-red marbles is \( ^{6}C_{3} \).
\( ^{6}C_{3} = \frac{6!}{3!3!} = 20 \).
Required probability \( = \frac{^{6}C_{3}}{^{12}C_{3}} = \frac{20}{220} = \frac{1}{11} \).
(v) Probability that not all of them are red:
This is the complement of "all of them are red."
\( P(\text{not all red}) = 1 - P(\text{all red}) \).
From part (ii), \( P(\text{all red}) = \frac{1}{11} \).
Required probability \( = 1 - \frac{1}{11} = \frac{10}{11} \).
(vi) Probability that none is black:
There are no black marbles in the bag. So, it's impossible to draw a black marble. This means any marble drawn will not be black.
The number of ways to draw 3 marbles, none of which are black, is the same as drawing 3 marbles from the total of 12 marbles (since all are non-black).
Number of favorable cases = \( ^{12}C_{3} = 220 \).
Required probability \( = \frac{220}{220} = 1 \). This event is certain to happen.
In simple words: From a bag of 12 marbles, we pick 3. First, we count all possible ways to pick 3 marbles. Then, for each question, we count how many ways match what we want (e.g., all blue). We divide these counts to find the chance. If it's impossible (like picking 3 white marbles when there are only 2), the chance is zero. If it's certain (like not picking a black marble when there are no black ones), the chance is one.

🎯 Exam Tip: "Not all red" is different from "none red." "Not all red" means at least one marble is not red, while "none red" means all marbles are not red. The complement rule \( P(A') = 1 - P(A) \) is very useful for "not all" scenarios.

Question 6. Two balls are drawn from an urn containing 2 white, 3 red and 4 black balls, one by one without replacement. What is the probability that (i) both balls are of the same colour, (ii) at least one ball is red ?
Answer: Total number of balls: \( 2 \text{ (white)} + 3 \text{ (red)} + 4 \text{ (black)} = 9 \) balls.
Balls are drawn one by one without replacement. This means the total number of items decreases after each draw.
(i) Probability that both balls are of the same colour:
This can happen in three ways: both are white (WW), both are red (RR), or both are black (BB).
\( P(\text{WW}) = P(\text{1st white}) \times P(\text{2nd white} | \text{1st white}) = \frac{2}{9} \times \frac{1}{8} = \frac{2}{72} \)
\( P(\text{RR}) = P(\text{1st red}) \times P(\text{2nd red} | \text{1st red}) = \frac{3}{9} \times \frac{2}{8} = \frac{6}{72} \)
\( P(\text{BB}) = P(\text{1st black}) \times P(\text{2nd black} | \text{1st black}) = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72} \)
Since these are mutually exclusive events, we add their probabilities:
\( P(\text{same colour}) = P(\text{WW}) + P(\text{RR}) + P(\text{BB}) = \frac{2}{72} + \frac{6}{72} + \frac{12}{72} = \frac{20}{72} = \frac{5}{18} \).
(ii) Probability that at least one ball is red:
"At least one red" means one red and one not-red, or both red.
It's easier to calculate the complement: \( 1 - P(\text{no red balls drawn}) \).
If no red balls are drawn, then both balls must be from the white or black balls.
Number of non-red balls = \( 2 \text{ (white)} + 4 \text{ (black)} = 6 \) balls.
\( P(\text{no red balls}) = P(\text{both from non-red}) = P(\text{1st non-red}) \times P(\text{2nd non-red} | \text{1st non-red}) = \frac{6}{9} \times \frac{5}{8} = \frac{30}{72} = \frac{5}{12} \).
Required probability \( = 1 - P(\text{no red balls}) = 1 - \frac{5}{12} = \frac{7}{12} \). The complement rule simplifies calculations significantly here.
In simple words: You have a pot with 9 balls of different colors. You take out two balls, one after the other, and don't put the first one back.
(i) We want to know the chance that both balls you picked are the same color. This means they are both white, or both red, or both black. We calculate the chance for each and add them up.
(ii) We want the chance that at least one of the balls is red. It's easier to find the chance that *neither* ball is red, and then subtract that from 1.

🎯 Exam Tip: When dealing with "at least one" probabilities, it's often more efficient to use the complement rule: \( P(\text{at least one}) = 1 - P(\text{none}) \).

Question 7. From a pack of cards, three are drawn at random. Find the chance that they are a king, a queen and a knave.
Answer: A standard pack of cards has 52 cards.
We are drawing three cards at random. The total number of ways to draw 3 cards from 52 is \( ^{52}C_{3} \).
\( ^{52}C_{3} = \frac{52!}{3!49!} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 52 \times 17 \times 25 = 22100 \). This is the total number of possible outcomes.
Now, we need to find the number of ways to draw one king, one queen, and one knave (Jack).
There are 4 kings in a deck, so we choose 1 king in \( ^{4}C_{1} = 4 \) ways.
There are 4 queens in a deck, so we choose 1 queen in \( ^{4}C_{1} = 4 \) ways.
There are 4 knaves (Jacks) in a deck, so we choose 1 knave in \( ^{4}C_{1} = 4 \) ways.
The total number of favorable cases (drawing one king, one queen, and one knave) is the product of these choices: \( ^{4}C_{1} \times ^{4}C_{1} \times ^{4}C_{1} = 4 \times 4 \times 4 = 64 \).
The required probability is the number of favorable outcomes divided by the total number of outcomes.
Required probability \( = \frac{64}{22100} = \frac{16}{5525} \). Dividing by 4 simplifies the fraction.
In simple words: We pick 3 cards from a full deck. We want to know the chance that these 3 cards are one King, one Queen, and one Jack. First, we count all the possible ways to pick any 3 cards. Then, we count all the ways to pick exactly one King, one Queen, and one Jack. Finally, we divide the second count by the first count to get the probability.

🎯 Exam Tip: When selecting distinct cards (like a king, a queen, and a knave), remember that each selection is independent, so you multiply the individual combination counts for each type of card.

Question 8.
(i) Two cards are drawn at random from 8 cards numbered from 1 to 8. What is the probability that the sum of the numbers is odd, if the two cards are drawn together ?
(ii) Two cards are drawn at random from a well shuffled pack of 52 cards, show that the chances of drawing two aces is \( \frac { 1 }{ 221 } \).
(iii) From a pack of 52 cards, 3 cards are drawn 'at random'. Find the probability of drawing exactly two aces.
(iv) Three cards are drawn at a time at random from a well shuffled pack of 52 cards. Find the probability that all the three cards have the same number.
(v) Two cards are drawn from a well shuffled pack of cards without replacement. Find the probability that neither a Jack nor a card of spades is drawn.
Answer:
(i) Two cards are drawn from 8 cards (numbered 1 to 8).
Total number of ways to draw 2 cards from 8 is \( ^{8}C_{2} \).
\( ^{8}C_{2} = \frac{8 \times 7}{2 \times 1} = 28 \).
For the sum of the numbers to be odd, one card must be odd and the other must be even.
The odd numbers from 1 to 8 are {1, 3, 5, 7} (4 odd numbers).
The even numbers from 1 to 8 are {2, 4, 6, 8} (4 even numbers).
Number of ways to choose 1 odd card from 4 is \( ^{4}C_{1} = 4 \).
Number of ways to choose 1 even card from 4 is \( ^{4}C_{1} = 4 \).
Number of favorable outcomes (1 odd and 1 even) is \( ^{4}C_{1} \times ^{4}C_{1} = 4 \times 4 = 16 \).
Required probability \( = \frac{16}{28} = \frac{4}{7} \).
(ii) Two cards are drawn from a pack of 52 cards.
Total number of ways to draw 2 cards from 52 is \( ^{52}C_{2} \).
\( ^{52}C_{2} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326 \).
There are 4 aces in a pack of 52 cards.
Number of ways to draw 2 aces from 4 is \( ^{4}C_{2} \).
\( ^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6 \).
Required probability \( = \frac{^{4}C_{2}}{^{52}C_{2}} = \frac{6}{1326} = \frac{1}{221} \). This matches the given value.
(iii) Three cards are drawn from a pack of 52 cards.
Total number of ways to draw 3 cards from 52 is \( ^{52}C_{3} \).
\( ^{52}C_{3} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100 \).
We want to draw exactly two aces.
Number of ways to choose 2 aces from 4 aces is \( ^{4}C_{2} = 6 \).
The third card must not be an ace. There are \( 52 - 4 = 48 \) non-ace cards.
Number of ways to choose 1 non-ace card from 48 is \( ^{48}C_{1} = 48 \).
Number of favorable outcomes (2 aces and 1 non-ace) = \( ^{4}C_{2} \times ^{48}C_{1} = 6 \times 48 = 288 \).
Required probability \( = \frac{288}{22100} = \frac{72}{5525} \).
(iv) Three cards are drawn from a pack of 52 cards.
Total number of ways to draw 3 cards from 52 is \( ^{52}C_{3} = 22100 \).
We want all three cards to have the same number. For example, three Kings or three 7s.
There are 13 different numbers/ranks in a deck (A, 2, 3, ..., 10, J, Q, K).
For each number/rank, there are 4 cards (one for each suit).
Number of ways to choose 3 cards of the same number from 4 cards (e.g., 3 Kings from 4 Kings) is \( ^{4}C_{3} = 4 \).
Since there are 13 different numbers/ranks, the total number of favorable outcomes is \( 13 \times ^{4}C_{3} = 13 \times 4 = 52 \).
Required probability \( = \frac{52}{22100} = \frac{1}{425} \).
(v) Two cards are drawn from a pack of 52 cards without replacement.
Total number of ways to draw 2 cards from 52 is \( ^{52}C_{2} = 1326 \).
We want neither a Jack nor a card of spades to be drawn.
Number of Jacks = 4. Number of spades = 13.
The Jack of spades is common to both.
Number of cards that are Jacks OR spades \( = (\text{number of Jacks}) + (\text{number of spades}) - (\text{number of Jack of spades}) \)
\( = 4 + 13 - 1 = 16 \).
Number of cards that are NEITHER a Jack NOR a spade = Total cards - (Jacks OR spades) \( = 52 - 16 = 36 \).
Number of ways to draw 2 cards from these 36 non-Jack and non-spade cards is \( ^{36}C_{2} \).
\( ^{36}C_{2} = \frac{36 \times 35}{2 \times 1} = 18 \times 35 = 630 \).
Required probability \( = \frac{^{36}C_{2}}{^{52}C_{2}} = \frac{630}{1326} = \frac{105}{221} \).
In simple words: This question asks for different probabilities when picking cards. We always start by finding the total ways to pick the cards. Then, for each part, we count the specific ways that match what's asked (like picking an odd and an even number, or two aces). We divide the specific ways by the total ways to get the probability. When cards are picked one after another without putting them back, the total number of cards decreases for the next pick.

🎯 Exam Tip: For "neither A nor B" probability, it's often easiest to find the number of elements that are in A OR B, and subtract this from the total. The remaining elements are neither A nor B.

Question 9. Three cards are drawn from a deck of 52 cards. What is the probability that
(i) they are all spades ?
(ii) they are all red cards ?
(iii) none of them is a club ?
(iv) all of them are aces ?
Answer: Three cards are drawn from a deck of 52 cards.
Total number of exhaustive outcomes = Total number of ways of drawing 3 cards out of 52 = \( ^{52}C_{3} \).
\( ^{52}C_{3} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} = 22100 \).
(i) Probability that they are all spades:
There are 13 spades in a deck.
Total number of favorable cases = Number of ways to draw 3 spades from 13 = \( ^{13}C_{3} \).
\( ^{13}C_{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 13 \times 2 \times 11 = 286 \).
Required probability \( = \frac{^{13}C_{3}}{^{52}C_{3}} = \frac{286}{22100} = \frac{11}{850} \).
(ii) Probability that they are all red cards:
There are 26 red cards in a deck (13 hearts + 13 diamonds).
Total number of favorable cases = Number of ways to draw 3 red cards from 26 = \( ^{26}C_{3} \).
\( ^{26}C_{3} = \frac{26 \times 25 \times 24}{3 \times 2 \times 1} = 26 \times 25 \times 4 = 2600 \).
Required probability \( = \frac{^{26}C_{3}}{^{52}C_{3}} = \frac{2600}{22100} = \frac{26}{221} = \frac{2}{17} \).
(iii) Probability that none of them is a club:
If none of the cards are clubs, they must be drawn from the other three suits (hearts, diamonds, spades).
Total cards that are not clubs = Total cards - Number of clubs \( = 52 - 13 = 39 \) cards.
Total number of favorable cases = Number of ways to draw 3 cards from these 39 non-club cards = \( ^{39}C_{3} \).
\( ^{39}C_{3} = \frac{39 \times 38 \times 37}{3 \times 2 \times 1} = 13 \times 19 \times 37 = 9139 \).
Required probability \( = \frac{^{39}C_{3}}{^{52}C_{3}} = \frac{9139}{22100} = \frac{703}{1700} \).
(iv) Probability that all of them are aces:
There are 4 aces in a deck.
Total number of favorable cases = Number of ways to draw 3 aces from 4 = \( ^{4}C_{3} \).
\( ^{4}C_{3} = \frac{4!}{3!1!} = 4 \).
Required probability \( = \frac{^{4}C_{3}}{^{52}C_{3}} = \frac{4}{22100} = \frac{1}{5525} \). It is very rare to draw three aces.
In simple words: We pick 3 cards from a deck.
(i) What's the chance all 3 are spades? Count how many ways to pick 3 spades and divide by all ways to pick 3 cards.
(ii) What's the chance all 3 are red? Count how many ways to pick 3 red cards and divide by all ways to pick 3 cards.
(iii) What's the chance none are clubs? Count how many ways to pick 3 cards that are *not* clubs, and divide by all ways to pick 3 cards.
(iv) What's the chance all 3 are aces? Count how many ways to pick 3 aces and divide by all ways to pick 3 cards.

🎯 Exam Tip: Always clearly identify the total number of items, the number of items in the specific category, and the number of items being chosen. Using the combination formula \( ^{n}C_{r} \) correctly is key.

Question 10. In Q. 7 what are the odds that
(i) all three cards are spades?
(ii) they are all red cards?
(iii) none is a club?
(iv) all of them are aces?
Answer: This question asks for "odds" instead of probability. Odds in favor of an event are given by \( P(\text{Event}) : P(\text{not Event}) \).
From Question 9, we already calculated the probabilities for these events. The total number of ways to draw 3 cards from 52 is \( ^{52}C_{3} = 22100 \).
(i) Odds that all three cards are spades:
From Q9(i), \( P(\text{all spades}) = \frac{11}{850} \).
\( P(\text{not all spades}) = 1 - P(\text{all spades}) = 1 - \frac{11}{850} = \frac{850 - 11}{850} = \frac{839}{850} \).
Odds in favor \( = P(\text{all spades}) : P(\text{not all spades}) = \frac{11}{850} : \frac{839}{850} = 11 : 839 \).
(ii) Odds that they are all red cards:
From Q9(ii), \( P(\text{all red cards}) = \frac{2}{17} \).
\( P(\text{not all red cards}) = 1 - P(\text{all red cards}) = 1 - \frac{2}{17} = \frac{17 - 2}{17} = \frac{15}{17} \).
Odds in favor \( = P(\text{all red cards}) : P(\text{not all red cards}) = \frac{2}{17} : \frac{15}{17} = 2 : 15 \).
(iii) Odds that none is a club:
From Q9(iii), \( P(\text{none is a club}) = \frac{703}{1700} \).
\( P(\text{at least one is a club}) = 1 - P(\text{none is a club}) = 1 - \frac{703}{1700} = \frac{1700 - 703}{1700} = \frac{997}{1700} \).
Odds in favor \( = P(\text{none is a club}) : P(\text{at least one is a club}) = \frac{703}{1700} : \frac{997}{1700} = 703 : 997 \).
(iv) Odds that all of them are aces:
From Q9(iv), \( P(\text{all aces}) = \frac{1}{5525} \).
\( P(\text{not all aces}) = 1 - P(\text{all aces}) = 1 - \frac{1}{5525} = \frac{5525 - 1}{5525} = \frac{5524}{5525} \).
Odds in favor \( = P(\text{all aces}) : P(\text{not all aces}) = \frac{1}{5525} : \frac{5524}{5525} = 1 : 5524 \). It's very unlikely.
In simple words: This asks for "odds" which is a way to compare the chance something happens to the chance it doesn't happen. We use the probabilities we found before. For example, if the chance of something is \( \frac{1}{5} \), then the chance it doesn't happen is \( \frac{4}{5} \). The odds would be \( 1:4 \). We just make sure to simplify the ratio.

🎯 Exam Tip: Always distinguish between "probability" and "odds." Probability is favorable outcomes divided by total outcomes. Odds in favor are favorable outcomes compared to unfavorable outcomes.

Question 11. One card is drawn from a pack of 52 cards, each of the 52 cards being equally like to be drawn. Find the probability that the card drawn is (i) an ace, (ii) red, (iii) either red or king, (iv) red and a king.
Answer: There are 52 cards in a standard deck. We are drawing one card.
(i) Probability that the card drawn is an ace:
There are 4 aces in a deck (Ace of Spades, Ace of Hearts, Ace of Diamonds, Ace of Clubs).
Required probability \( = \frac{\text{Number of aces}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13} \).
(ii) Probability that the card drawn is red:
There are 26 red cards in a deck (13 hearts and 13 diamonds).
Required probability \( = \frac{\text{Number of red cards}}{\text{Total number of cards}} = \frac{26}{52} = \frac{1}{2} \).
(iii) Probability that the card drawn is either red or a king:
Let A be the event of drawing a red card, and B be the event of drawing a king.
We need to find \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
\( P(A) = P(\text{red}) = \frac{26}{52} \).
\( P(B) = P(\text{king}) = \frac{4}{52} \).
\( P(A \cap B) = P(\text{red and king}) \). There are 2 red kings (King of Hearts and King of Diamonds).
\( P(A \cap B) = \frac{2}{52} \).
Required probability \( = \frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{26 + 4 - 2}{52} = \frac{28}{52} = \frac{7}{13} \).
(iv) Probability that the card drawn is red and a king:
This means the card must be one of the red kings.
There are 2 red kings in a deck (King of Hearts and King of Diamonds).
Required probability \( = \frac{\text{Number of red kings}}{\text{Total number of cards}} = \frac{2}{52} = \frac{1}{26} \).
In simple words: When you pick just one card from a deck, we look at its chance of being different types.
(i) For an Ace, we count the 4 Aces and divide by 52.
(ii) For a Red card, we count the 26 Red cards and divide by 52.
(iii) For a Red card OR a King, we add the number of Red cards and Kings, but we must subtract the Kings that are also Red (because we counted them twice). Then divide by 52.
(iv) For a Red card AND a King, we only count the cards that are both Red and Kings (King of Hearts and King of Diamonds) and divide by 52.

🎯 Exam Tip: Remember the formula for the probability of A or B: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). This avoids double-counting elements that belong to both categories.

Question 12. Four cards are drawn at random from a pack of 52 playing cards, Find the probability of getting
(i) all the four cards of the same suit;
(ii) all the four cards of the same number;
(iii) one card from each suit;
(iv) two red cards and two black cards;
(v) all cards of the same colour ;
(vi) all face cards ; (King, Queen, Jack)
(vii) four honours of the same suit.
Answer: Four cards are drawn from a pack of 52 playing cards.
Total number of exhaustive cases = Total number of ways of drawing 4 cards out of 52 = \( ^{52}C_{4} \).
\( ^{52}C_{4} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 13 \times 17 \times 25 \times 49 = 270725 \).
(i) Probability of getting all four cards of the same suit:
There are 4 suits, and each suit has 13 cards.
Number of ways to choose 4 cards from a specific suit (e.g., all 4 spades) is \( ^{13}C_{4} \).
\( ^{13}C_{4} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 13 \times 11 \times 5 = 715 \).
Since there are 4 suits, the total number of favorable outcomes is \( 4 \times ^{13}C_{4} = 4 \times 715 = 2860 \).
Required probability \( = \frac{2860}{270725} = \frac{13}{1225} \). (Dividing by 220)
(ii) Probability of getting all four cards of the same number:
There are 13 different numbers/ranks (A, 2, ..., K).
For each number, there are 4 cards (one for each suit).
Number of ways to choose 4 cards of the same number from 4 available cards (e.g., all 4 Aces) is \( ^{4}C_{4} = 1 \).
Since there are 13 different numbers/ranks, the total number of favorable outcomes is \( 13 \times ^{4}C_{4} = 13 \times 1 = 13 \).
Required probability \( = \frac{13}{270725} \).
(iii) Probability of getting one card from each suit:
There are 4 suits, each with 13 cards. We need to choose 1 card from each suit.
Number of ways to choose 1 card from 13 spades is \( ^{13}C_{1} = 13 \).
Number of ways to choose 1 card from 13 hearts is \( ^{13}C_{1} = 13 \).
Number of ways to choose 1 card from 13 diamonds is \( ^{13}C_{1} = 13 \).
Number of ways to choose 1 card from 13 clubs is \( ^{13}C_{1} = 13 \).
Total number of favorable outcomes \( = ^{13}C_{1} \times ^{13}C_{1} \times ^{13}C_{1} \times ^{13}C_{1} = 13 \times 13 \times 13 \times 13 = 13^4 = 28561 \).
Required probability \( = \frac{28561}{270725} = \frac{2197}{20825} \). (Dividing by 13)
(iv) Probability of getting two red cards and two black cards:
There are 26 red cards and 26 black cards in a deck.
Number of ways to choose 2 red cards from 26 is \( ^{26}C_{2} \).
\( ^{26}C_{2} = \frac{26 \times 25}{2 \times 1} = 13 \times 25 = 325 \).
Number of ways to choose 2 black cards from 26 is \( ^{26}C_{2} = 325 \).
Total number of favorable outcomes \( = ^{26}C_{2} \times ^{26}C_{2} = 325 \times 325 = 105625 \).
Required probability \( = \frac{105625}{270725} = \frac{325}{833} \).
(v) Probability of getting all cards of the same colour:
This means all 4 cards are either red OR all 4 cards are black.
Number of ways to choose 4 red cards from 26 is \( ^{26}C_{4} \).
\( ^{26}C_{4} = \frac{26 \times 25 \times 24 \times 23}{4 \times 3 \times 2 \times 1} = 26 \times 25 \times 23 = 14950 \).
Number of ways to choose 4 black cards from 26 is \( ^{26}C_{4} = 14950 \).
Total number of favorable outcomes = \( ^{26}C_{4} + ^{26}C_{4} = 14950 + 14950 = 29900 \).
Required probability \( = \frac{29900}{270725} = \frac{116}{1063} \). (Dividing by 25)
(vi) Probability of getting all face cards (King, Queen, Jack):
There are 12 face cards in a deck (4 Kings + 4 Queens + 4 Jacks).
Total number of favorable cases = Number of ways to choose 4 face cards from 12 = \( ^{12}C_{4} \).
\( ^{12}C_{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 11 \times 5 \times 9 = 495 \).
Required probability \( = \frac{495}{270725} = \frac{99}{54145} \).
(vii) Probability of getting four honours of the same suit:
Honour cards are Ace, King, Queen, Jack, 10. In each suit, there are 5 honour cards.
If the question means picking 4 of these 5 cards from the *same* suit, then:
Number of ways to pick 4 honour cards from 5 in a single suit is \( ^{5}C_{4} = 5 \).
Since there are 4 suits, total favorable outcomes = \( 4 \times ^{5}C_{4} = 4 \times 5 = 20 \).
Required probability \( = \frac{20}{270725} = \frac{4}{54145} \). (Dividing by 5)
*Note: Sometimes "honour cards" can refer to A, K, Q, J only, depending on context. Assuming A, K, Q, J, 10 here as it's common in Bridge/Poker point counting, and the source calculation leads to this interpretation.*
In simple words: We are picking 4 cards from a full deck.
(i) To get all cards of the same suit (like 4 spades), we count how many ways to pick 4 from one suit and multiply by 4 (for the 4 suits).
(ii) To get all cards of the same number (like 4 Aces), we count how many ranks there are (13) and for each rank, there's only 1 way to pick all 4 cards of that rank.
(iii) To get one card from each suit, we pick one from spades, one from hearts, one from diamonds, and one from clubs, and multiply the possibilities.
(iv) For two red and two black cards, we pick 2 from the 26 red cards and 2 from the 26 black cards, then multiply these choices.
(v) For all cards of the same color, we find the ways to pick 4 red cards OR 4 black cards and add them.
(vi) For all face cards (Kings, Queens, Jacks), we count how many face cards there are (12) and pick 4 from them.
(vii) For four honour cards of the same suit, we choose 4 from the 5 honour cards (A, K, Q, J, 10) in one suit, and multiply by 4 for the 4 suits. We always divide by the total ways to pick 4 cards from the deck.

🎯 Exam Tip: Pay close attention to keywords like "same suit," "same number," "each suit," "same color," and specific card types. Each requires a different approach to calculate favorable outcomes.

Question 13. In a lottery of 50 tickets numbered 1 to 50, two tickets are drawn simultaneously. Find the probability that
(i) both the tickets drawn have prime numbers ;
(ii) none of the tickets drawn has prime number ;
(iii) a ticket has prime number.
Answer: Total number of tickets = 50.
Two tickets are drawn simultaneously.
Total number of exhaustive cases = Total number of ways to draw 2 tickets from 50 = \( ^{50}C_{2} \).
\( ^{50}C_{2} = \frac{50 \times 49}{2 \times 1} = 25 \times 49 = 1225 \).
(i) Probability that both tickets drawn have prime numbers:
First, list the prime numbers between 1 and 50: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}.
There are 15 prime numbers.
Total number of favorable cases = Number of ways to draw 2 prime tickets from these 15 prime numbers = \( ^{15}C_{2} \).
\( ^{15}C_{2} = \frac{15 \times 14}{2 \times 1} = 15 \times 7 = 105 \).
Required probability \( = \frac{105}{1225} = \frac{3 \times 35}{35 \times 35} = \frac{3}{35} \).
(ii) Probability that none of the tickets drawn has a prime number:
This means both tickets drawn must be non-prime numbers.
Number of non-prime tickets = Total tickets - Number of prime tickets \( = 50 - 15 = 35 \).
Total number of favorable cases = Number of ways to draw 2 non-prime tickets from these 35 non-prime numbers = \( ^{35}C_{2} \).
\( ^{35}C_{2} = \frac{35 \times 34}{2 \times 1} = 35 \times 17 = 595 \).
Required probability \( = \frac{595}{1225} = \frac{17 \times 35}{35 \times 35} = \frac{17}{35} \).
(iii) Probability that a ticket has a prime number (This phrasing often means "exactly one ticket has a prime number"):
This means one ticket is prime, and the other ticket is non-prime.
Number of ways to choose 1 prime ticket from 15 is \( ^{15}C_{1} = 15 \).
Number of ways to choose 1 non-prime ticket from 35 is \( ^{35}C_{1} = 35 \).
Total number of favorable cases = \( ^{15}C_{1} \times ^{35}C_{1} = 15 \times 35 = 525 \).
Required probability \( = \frac{525}{1225} = \frac{15 \times 35}{35 \times 35} = \frac{15}{35} = \frac{3}{7} \).
In simple words: From 50 numbered tickets, two are picked.
(i) What's the chance both are prime numbers? First, list all prime numbers up to 50. Then count ways to pick 2 from these primes, and divide by all ways to pick 2 tickets.
(ii) What's the chance neither is a prime number? Count numbers that are not prime. Then count ways to pick 2 from these non-primes, and divide by all ways to pick 2 tickets.
(iii) What's the chance exactly one ticket is prime? This means one is prime AND the other is not prime. Multiply the ways to pick one prime by the ways to pick one non-prime. Then divide by all ways to pick 2 tickets.

🎯 Exam Tip: Be careful with the phrasing "a ticket has prime number." In probability, this usually implies "exactly one." If it meant "at least one," the calculation would be \( 1 - P(\text{none are prime}) \).

Question 14.
(i) Out of 9 outstanding students in a college, there are 4 boys and 5 girls. A team of 4 students is to be selected for a quiz programme. Find the probability that 2 are girls and 2 are boys.
(ii) Four people are chosen at random from a group consisting of 3 men, 2 women, and 3 children. Find the probability that out of four chosen people, exactly 2 are children ?
Answer:
(i) Total number of students = 9 (4 boys + 5 girls).
A team of 4 students is to be selected.
Total number of ways to select 4 students from 9 is \( ^{9}C_{4} \).
\( ^{9}C_{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 9 \times 2 \times 7 = 126 \).
We want the probability that the team has 2 girls and 2 boys.
Number of ways to select 2 girls from 5 is \( ^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10 \).
Number of ways to select 2 boys from 4 is \( ^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6 \).
Total number of favorable cases (2 girls and 2 boys) = \( ^{5}C_{2} \times ^{4}C_{2} = 10 \times 6 = 60 \).
Required probability \( = \frac{60}{126} = \frac{10}{21} \). (Dividing by 6)
(ii) Total number of people in the group = 3 men + 2 women + 3 children = 8 people.
Four people are chosen at random.
Total number of ways to choose 4 people from 8 is \( ^{8}C_{4} \).
\( ^{8}C_{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \).
We want the probability that exactly 2 of the four chosen people are children.
Number of ways to choose 2 children from 3 is \( ^{3}C_{2} = \frac{3 \times 2}{2 \times 1} = 3 \).
The remaining \( 4 - 2 = 2 \) people must be chosen from the non-children (men and women).
Number of non-children = 3 men + 2 women = 5 people.
Number of ways to choose 2 non-children from 5 is \( ^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10 \).
Total number of favorable cases (2 children and 2 non-children) = \( ^{3}C_{2} \times ^{5}C_{2} = 3 \times 10 = 30 \).
Required probability \( = \frac{30}{70} = \frac{3}{7} \). It is important to remember to include all remaining categories for selection.
In simple words:
(i) From 9 students (4 boys, 5 girls), we pick a team of 4. We want to know the chance of picking 2 girls and 2 boys. We count how many ways to pick 2 girls and how many ways to pick 2 boys, multiply them, then divide by all ways to pick 4 students.
(ii) From a group of 8 people (3 men, 2 women, 3 children), we pick 4. We want to know the chance of picking exactly 2 children. We count ways to pick 2 children, then ways to pick the other 2 people from the non-children (men and women). Multiply these and divide by all ways to pick 4 people.

🎯 Exam Tip: When selecting a subgroup with specific compositions (e.g., 2 girls and 2 boys), calculate the combinations for each part of the composition and then multiply them. The denominator is always the total combination of selecting the full subgroup.

Question 15. A committee of 5 principals is to be selected from a group of 6 grant principals and 8 lady principals. If the selected is made randomly, find the probability that there are 3 lady principals and 2 gent principals.
Answer: Total number of principals = 6 grant principals + 8 lady principals = 14 principals.
A committee of 5 principals is to be selected.
Total number of ways to select 5 principals from 14 is \( ^{14}C_{5} \).
\( ^{14}C_{5} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 14 \times 13 \times 11 = 2002 \).
We want the probability that the committee has 3 lady principals and 2 grant (gent) principals.
Number of ways to select 3 lady principals from 8 is \( ^{8}C_{3} \).
\( ^{8}C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \).
Number of ways to select 2 grant principals from 6 is \( ^{6}C_{2} \).
\( ^{6}C_{2} = \frac{6 \times 5}{2 \times 1} = 15 \).
Total number of favorable cases (3 lady principals and 2 grant principals) = \( ^{8}C_{3} \times ^{6}C_{2} = 56 \times 15 = 840 \).
Required probability \( = \frac{840}{2002} = \frac{60}{143} \). (Dividing by 14)
In simple words: We have 14 principals (6 grant, 8 lady). We need to pick a committee of 5. We want to find the chance that this committee has exactly 3 lady principals and 2 grant principals. First, we count all possible ways to pick any 5 principals. Then, we count how many ways we can pick 3 ladies and 2 grant principals. Finally, we divide these two counts.

🎯 Exam Tip: Always clearly list the different categories and their counts. When forming a mixed group, multiply the combinations for each category to find the total favorable outcomes.

Question 16. A bag contains tickets numbered 1 to 20. Two tickets are drawn. Find the probability that both numbers are odd.
Answer: Total number of tickets = 20 (numbered 1 to 20).
Two tickets are drawn.
Total number of ways to draw 2 tickets from 20 is \( ^{20}C_{2} \).
\( ^{20}C_{2} = \frac{20 \times 19}{2 \times 1} = 10 \times 19 = 190 \).
Now, we need to find the number of odd tickets between 1 and 20.
Odd numbers are {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}.
There are 10 odd numbers.
We want the probability that both drawn tickets have odd numbers.
Total number of favorable cases = Number of ways to draw 2 odd tickets from these 10 odd numbers = \( ^{10}C_{2} \).
\( ^{10}C_{2} = \frac{10 \times 9}{2 \times 1} = 5 \times 9 = 45 \).
Required probability \( = \frac{45}{190} = \frac{9}{38} \). (Dividing by 5) This means it's less than a quarter chance.
In simple words: You have 20 tickets, numbered 1 to 20. You pick two tickets. We want to know the chance that both tickets have odd numbers. First, count all the odd numbers between 1 and 20. Then, count all the ways you can pick two odd tickets. Divide this by all the ways you can pick any two tickets from the 20.

🎯 Exam Tip: When dealing with numbered items, correctly identifying the count of items that fit a specific property (like odd/even, prime/composite) is the first critical step.

Question 17. A bag contains tickets numbered 1 to 30. Three tickets are drawn at random from the bag. What is the probability that the maximum number of the selected tickets exceeds 25 ?
Answer: Total number of tickets = 30 (numbered 1 to 30).
Three tickets are drawn at random.
Total number of ways to draw 3 tickets from 30 is \( ^{30}C_{3} \).
\( ^{30}C_{3} = \frac{30 \times 29 \times 28}{3 \times 2 \times 1} = 10 \times 29 \times 14 = 4060 \).
Let A be the event that the maximum number of the selected tickets exceeds 25.
This means at least one of the three tickets drawn is 26, 27, 28, 29, or 30.
It is easier to calculate the complement event, \( A' \), which is that the maximum number of the selected tickets does NOT exceed 25. This means all three tickets drawn must be 25 or less.
Number of tickets that are 25 or less = 25 (i.e., tickets numbered 1 to 25).
Number of ways to draw 3 tickets from these 25 tickets is \( ^{25}C_{3} \).
\( ^{25}C_{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 25 \times 4 \times 23 = 2300 \).
The probability of event \( A' \) (none of the tickets exceed 25) is:
\( P(A') = \frac{^{25}C_{3}}{^{30}C_{3}} = \frac{2300}{4060} = \frac{230}{406} = \frac{115}{203} \).
The required probability for event A is \( P(A) = 1 - P(A') \).
\( P(A) = 1 - \frac{115}{203} = \frac{203 - 115}{203} = \frac{88}{203} \). The complement rule simplified this greatly.
In simple words: You pick 3 tickets from 30. We want to find the chance that the highest number on any of your 3 tickets is bigger than 25. It's easier to find the chance that *none* of your tickets have a number higher than 25 (meaning all 3 tickets are 25 or less). Then, you subtract this chance from 1 to get the answer.

🎯 Exam Tip: For "maximum number exceeds X" or "at least one item has property Y," using the complement rule (1 - P(none have property Y)) is almost always simpler and less prone to errors.

Question 18.
(i) A room has 3 lamps. From a collection of 10 light bulbs of which 6 are no good, a person selects 3 at random and puts them in a socket. What is the probability, that he will have light ?
(ii) A has 3 shares in a lottery containing 3 prizes and 9 blanks ; B has two shares in lottery containing 2 prizes and 6 blanks. Compare their chances of success.
Answer:
(i) Total number of light bulbs = 10.
Number of no good (defective) bulbs = 6.
Number of good bulbs = Total bulbs - Defective bulbs \( = 10 - 6 = 4 \).
A person selects 3 bulbs at random. A room has 3 lamps, meaning if any of the three selected bulbs are good, there is light. The problem asks for the probability that "he will have light," which means at least one of the three bulbs selected must be good.
Total number of ways to select 3 bulbs from 10 is \( ^{10}C_{3} \).
\( ^{10}C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120 \).
It's easier to calculate the complement event: no light, which means all 3 selected bulbs are defective (no good).
Number of ways to select 3 defective bulbs from 6 defective bulbs is \( ^{6}C_{3} \).
\( ^{6}C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \).
Probability of no light \( = P(\text{all 3 defective}) = \frac{20}{120} = \frac{1}{6} \).
Required probability of having light \( = P(\text{at least one good bulb}) = 1 - P(\text{no light}) = 1 - \frac{1}{6} = \frac{5}{6} \). It is quite likely to have light.
(ii) Compare chances of success for A and B:
**For A:**
Total shares (tickets) for A's lottery = 3 prizes + 9 blanks = 12 shares.
A has 3 shares.
Probability of A not winning (i.e., all 3 shares are blanks) = \( \frac{^{9}C_{3}}{^{12}C_{3}} \).
\( ^{9}C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84 \).
\( ^{12}C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220 \).
\( P(\text{A not winning}) = \frac{84}{220} = \frac{21}{55} \).
Probability of A winning \( = 1 - P(\text{A not winning}) = 1 - \frac{21}{55} = \frac{34}{55} \).
**For B:**
Total shares (tickets) for B's lottery = 2 prizes + 6 blanks = 8 shares.
B has 2 shares.
Probability of B not winning (i.e., both 2 shares are blanks) = \( \frac{^{6}C_{2}}{^{8}C_{2}} \).
\( ^{6}C_{2} = \frac{6 \times 5}{2 \times 1} = 15 \).
\( ^{8}C_{2} = \frac{8 \times 7}{2 \times 1} = 28 \).
\( P(\text{B not winning}) = \frac{15}{28} \).
Probability of B winning \( = 1 - P(\text{B not winning}) = 1 - \frac{15}{28} = \frac{13}{28} \).
**Comparison:**
A's winning probability = \( \frac{34}{55} \).
B's winning probability = \( \frac{13}{28} \).
To compare, we can find a common denominator or convert to decimals.
\( \frac{34}{55} \approx 0.618 \)
\( \frac{13}{28} \approx 0.464 \)
So, A has a greater chance of success than B.
Alternatively, compare \( 34 \times 28 \) with \( 13 \times 55 \):
\( 34 \times 28 = 952 \)
\( 13 \times 55 = 715 \)
Since \( 952 > 715 \), A's probability \( \frac{34}{55} \) is greater than B's probability \( \frac{13}{28} \).
In simple words:
(i) You pick 3 light bulbs from 10, knowing 6 are broken and 4 are good. To have light, at least one bulb must work. It's easier to find the chance that *all three* are broken, then subtract that from 1.
(ii) Two people, A and B, play different lotteries. We find A's chance of winning by figuring out the chance A *doesn't* win and subtracting from 1. We do the same for B. Then we compare who has a better chance by looking at the numbers.

🎯 Exam Tip: "At least one" problems are prime candidates for using the complement rule. When comparing probabilities, convert to decimals or find a common denominator for accurate comparison.

Question 19.
(i) There are n letters and n addressed envelopes. If the letters are placed in the envelopes at random, what is the probability that all the letters are not placed in the right envelope ?
(ii) Three letters are written to different persons, and the addresses on the three envelopes are also written. Without looking at the addresses, find the probability that the letters go into the right envelopes.
Answer:
(i) First, let's find the total number of ways to place \( n \) letters into \( n \) addressed envelopes. This is given by \( n! \) (n factorial). Next, there is only one way for all \( n \) letters to be placed into their correct envelopes. So, the probability that all letters are placed correctly is \( \frac{1}{n!} \). Therefore, the probability that all letters are *not* placed in the right envelope is \( 1 - \frac{1}{n!} \). This is because for every way some letters are correctly placed, there are many ways others are mismatched.
(ii) For three letters and three envelopes, the total number of ways to place them is \( 3! \), which equals \( 3 \times 2 \times 1 = 6 \). There is only one way for all three letters to be placed into their correct envelopes. This fundamental principle of permutation helps calculate the likelihood of perfect matches in small sets. So, the probability that all letters are placed in the right envelopes is \( \frac{1}{6} \).
In simple words: For part (i), we find the chance that none of the letters go into their correct envelopes by taking 1 minus the chance that all of them do. For part (ii), with 3 letters and 3 envelopes, there are 6 ways to place them, and only 1 way for all to be correct.

🎯 Exam Tip: For probability problems involving permutations, clearly identify the total number of possible arrangements (exhaustive cases) and the number of desired arrangements (favourable cases).

Question 20. The letters of word 'SOCIETY' are placed at random in a row. What is the probability that three vowels come together ?
Answer: The word SOCIETY has 7 letters in total. The vowels are O, I, E (3 vowels), and the consonants are S, C, T, Y (4 consonants). The total number of ways to arrange all 7 letters in a row is \( 7! \). To find the probability that the three vowels come together, we treat the group of vowels (OIE) as a single unit. Now we have 1 vowel unit and 4 consonants, making a total of 5 units (OIE, S, C, T, Y). These 5 units can be arranged in \( 5! \) ways. Additionally, the 3 vowels within their own unit (OIE) can be arranged among themselves in \( 3! \) ways. Grouping similar elements is a common technique in permutation problems to simplify calculations. So, the total number of favourable arrangements where vowels come together is \( 5! \times 3! \). The required probability is the ratio of favourable arrangements to the total arrangements: \( P(\text{vowels together}) = \frac{5! \times 3!}{7!} = \frac{(5 \times 4 \times 3 \times 2 \times 1) \times (3 \times 2 \times 1)}{(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \) \( = \frac{120 \times 6}{5040} = \frac{720}{5040} = \frac{1}{7} \)
In simple words: The word SOCIETY has 7 letters. If we want the 3 vowels (O, I, E) to stick together, we treat them as one block. Then we arrange this block with the other 4 letters. Also, the vowels inside their block can move around. We divide the total ways they can stick together by all possible ways to arrange the letters.

🎯 Exam Tip: Remember to count the internal arrangements of the grouped elements when calculating favourable outcomes in permutation problems.

Question 21. In a random arrangement of the letters of the word "COMMERCE", find the probability that all vowels come together.
Answer: The word COMMERCE has 8 letters. The repeated letters are: C (2 times), M (2 times), E (2 times). There is one O and one R. The total number of distinct ways to arrange the letters of COMMERCE is \( \frac{8!}{2!2!2!} \). Now, let's identify the vowels: O, E, E (3 vowels). The consonants are C, C, M, M, R (5 consonants). To find the probability that all vowels come together, we treat the group of vowels (OEE) as a single unit. Now we have 1 vowel unit and 5 consonants. The 5 consonants are C, C, M, M, R. So, we are arranging 6 units (the vowel block and 5 consonants), where C is repeated 2 times and M is repeated 2 times. The number of ways to arrange these 6 units is \( \frac{6!}{2!2!} \). Within the vowel block (OEE), the 3 vowels can be arranged in \( \frac{3!}{2!} \) ways because E is repeated 2 times. So, the total number of favourable arrangements where all vowels come together is \( \frac{6!}{2!2!} \times \frac{3!}{2!} \). When dealing with repeated letters, remember to divide by the factorial of the count of each repeated letter. The required probability is: \( P(\text{vowels together}) = \frac{\frac{6!}{2!2!} \times \frac{3!}{2!}}{\frac{8!}{2!2!2!}} \) \( = \frac{6! \times 3! \times 2! \times 2!}{8! \times 2! \times 2! \times 2!} \) (After simplifying the factorials in the denominators) \( = \frac{6! \times 3!}{8!} \) \( = \frac{(6 \times 5 \times 4 \times 3 \times 2 \times 1) \times (3 \times 2 \times 1)}{(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \) \( = \frac{720 \times 6}{40320} = \frac{4320}{40320} = \frac{3}{28} \)
In simple words: For the word COMMERCE, we first find all possible ways to arrange its letters, remembering some are repeated. Then, we find ways where the vowels (OEE) stick together as one group, again considering repeated letters within that group and in the consonants. Finally, we divide the "vowels together" ways by the "all possible" ways.

🎯 Exam Tip: For words with repeated letters, always use the formula \( \frac{n!}{p_1! p_2! ...} \) for permutations, where \( n \) is the total number of letters and \( p_i \) are the counts of repeated letters.

Question 22. Given a group of 4 persons, find the probability that
(i) no two of them have their birthdays on the same day,
(ii) all of them have same birthday. [Ignore the existence of a leap year]
Answer: (We assume there are 365 days in a year, ignoring leap years.)
(i) **No two of them have their birthdays on the same day:**
The total number of possible ways for 4 people to have birthdays is \( 365 \times 365 \times 365 \times 365 = (365)^4 \), since each person can have a birthday on any of the 365 days. For no two persons to have the same birthday, the first person can have a birthday on any of the 365 days. The second person must have a birthday on one of the remaining 364 days. The third person must choose from 363 days, and the fourth person from 362 days. So, the number of favourable ways is \( 365 \times 364 \times 363 \times 362 \). The required probability is: \( P(\text{no shared birthdays}) = \frac{365 \times 364 \times 363 \times 362}{(365)^4} \) This can also be written using permutation notation as \( \frac{P(365, 4)}{365^4} \). This classic "birthday problem" shows how surprisingly quickly the probability of shared birthdays increases with more people.
(ii) **All of them have the same birthday:**
The total number of possible ways for 4 people to have birthdays is still \( (365)^4 \). For all four persons to have the same birthday, they must all share one specific day. There are 365 possible days they could all share (e.g., all born on January 1st, or all on January 2nd, etc.). So, the number of favourable ways is 365. The required probability is: \( P(\text{all same birthday}) = \frac{365}{(365)^4} = \frac{1}{(365)^3} \) This scenario is much less likely than even a single shared birthday among a group.
In simple words: For part (i), we want everyone to have a different birthday. The first person has all 365 days, the next has 364, and so on. We divide this by the total ways they could have birthdays. For part (ii), we want all four people to have the exact same birthday. There are 365 possible days they could share, and we divide that by the total ways they could have birthdays.

🎯 Exam Tip: When dealing with "at least one" or "no two" scenarios, consider using the complement rule (1 - P(none)) if it simplifies calculations. Remember to clarify if repetition is allowed or not, which changes the total possible outcomes significantly.