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

 

Question 1. What is the probability of getting :

Answer:
(i) Probability of getting a head on a coin: Total number of outcomes when flipping a coin is 2 (Head or Tail). Number of favourable outcomes (getting a Head) is 1. So, the required probability of getting a head is \( \frac{1}{2} \).
(ii) Probability of getting a 6 on a die: When a die is rolled, the total possible outcomes are {1, 2, 3, 4, 5, 6}, so there are 6 outcomes. The event of getting a 6 has only 1 favourable outcome. Thus, the required probability of getting a 6 is \( \frac{1}{6} \).
(iii) Probability of getting a prime number on the spinner: The spinner has numbers {1, 2, 3, 4, 5, 6, 7}. So, the total number of outcomes is 7. The prime numbers on the spinner are {2, 3, 5, 7}, which means there are 4 favourable outcomes. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Therefore, the required probability of getting a prime number is \( \frac{4}{7} \).
In simple words: To find probability, we count how many ways the event we want can happen and divide that by the total number of all possible outcomes. For example, a coin has 2 sides, and only 1 is a head, so the chance is 1 out of 2.

🎯 Exam Tip: Always clearly define the total sample space (n(S)) and the number of favourable outcomes (n(E)) for each part of a probability question to avoid errors.

 

Question 2. Ramesh chooses a date at random in April for a party. Calculate the probability that he chooses:
(i) a Saturday
(ii) a Sunday
(iii) a Saturday or a Sunday.

Answer:
From the given April calendar: Total number of days in April = 30. So, \( n(S) = 30 \).
(i) Let E be the event that Ramesh chooses a Saturday. The Saturdays in April are: {3, 10, 17, 24}. So, the number of Saturdays, \( n(E) = 4 \). The required probability \( = \frac{n(E)}{n(S)} = \frac{4}{30} = \frac{2}{15} \).
(ii) Let F be the event that Ramesh chooses a Sunday. The Sundays in April are: {4, 11, 18, 25}. So, the number of Sundays, \( n(F) = 4 \). The required probability \( = \frac{n(F)}{n(S)} = \frac{4}{30} = \frac{2}{15} \).
(iii) Let G be the event that Ramesh chooses a Saturday or a Sunday. The days that are a Saturday or a Sunday are: {3, 4, 10, 11, 17, 18, 24, 25}. So, the number of days, \( n(G) = 8 \). The required probability \( = \frac{n(G)}{n(S)} = \frac{8}{30} = \frac{4}{15} \).
In simple words: First, count all the days in April. Then, count how many of those days are Saturdays, Sundays, or both. Finally, divide the count of desired days by the total days in the month to find the probability.

🎯 Exam Tip: When dealing with calendar problems, carefully list out all the dates for the specified days (e.g., Saturdays, Sundays) to ensure accurate counting of favourable outcomes.

 

Question 3. A normal die is rolled. Calculate the probability that the number on the uppermost face when it stops rolling will be
(i) 5
(ii) not 5
(iii) an odd number
(iv) a prime number
(v) a 3 or a 4
(vi) a 1 or a 2 or a 3 or a 4.

Answer:
When a die is rolled, the sample space (S) consists of the numbers {1, 2, 3, 4, 5, 6}. The total number of outcomes, \( n(S) = 6 \).
(i) Let A be the event of getting a 5. The favourable outcome is {5}. So, \( n(A) = 1 \). The required probability \( = \frac{n(A)}{n(S)} = \frac{1}{6} \).
(ii) Let B be the event of getting not 5. The favourable outcomes are {1, 2, 3, 4, 6}. So, \( n(B) = 5 \). The required probability \( = \frac{n(B)}{n(S)} = \frac{5}{6} \).
(iii) Let C be the event of getting an odd number. The favourable outcomes are {1, 3, 5}. So, \( n(C) = 3 \). The required probability \( = \frac{n(C)}{n(S)} = \frac{3}{6} = \frac{1}{2} \).
(iv) Let D be the event of getting a prime number. The prime numbers are {2, 3, 5}. So, \( n(D) = 3 \). Prime numbers are only divisible by 1 and themselves. The required probability \( = \frac{n(D)}{n(S)} = \frac{3}{6} = \frac{1}{2} \).
(v) Let E be the event of getting a 3 or a 4. The favourable outcomes are {3, 4}. So, \( n(E) = 2 \). The required probability \( = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3} \).
(vi) Let F be the event of getting a 1 or a 2 or a 3 or a 4. The favourable outcomes are {1, 2, 3, 4}. So, \( n(F) = 4 \). The required probability \( = \frac{n(F)}{n(S)} = \frac{4}{6} = \frac{2}{3} \).
In simple words: When you roll a normal die, there are six possible outcomes. To find the chance of a specific event, count how many of those outcomes match your event and divide by six. For example, there are three odd numbers (1, 3, 5) out of six total, so the probability is 3/6 or 1/2.

🎯 Exam Tip: Remember to list the full sample space (S) for a die roll first, then identify the specific outcomes for each event to ensure correct counting and calculation of probabilities.

 

Question 4. Nine playing cards are numbered 2 to 10. A card is selected from them at random. Calculate the probability that the card will be :
(i) an odd number
(ii) a multiple of 4.

Answer:
The set of playing cards available is S = {2, 3, 4, 5, 6, 7, 8, 9, 10}. The total number of possible outcomes, \( n(S) = 9 \).
(i) Let E be the event of getting an odd-numbered card. The odd numbers in the set are {3, 5, 7, 9}. So, the number of favourable outcomes, \( n(E) = 4 \). The required probability \( = \frac{n(E)}{n(S)} = \frac{4}{9} \).
(ii) Let F be the event of getting a multiple of 4. The multiples of 4 in the set are {4, 8}. So, the number of favourable outcomes, \( n(F) = 2 \). The required probability \( = \frac{n(F)}{n(S)} = \frac{2}{9} \).
In simple words: If you have cards numbered 2 to 10, there are 9 cards in total. To find the chance of picking an odd number, count all the odd numbers (3, 5, 7, 9), which is 4, so it's 4 out of 9. For a multiple of 4, count those (4, 8), which is 2, so it's 2 out of 9.

🎯 Exam Tip: Always list out the numbers in your sample space before identifying events. This helps ensure you don't miss any outcomes or include numbers outside the given range.

 

Question 5. Nine counters numbered 2 to 10 are put in a bag. One counter is selected at random. What is the probability of getting a counter with :
(i) a number 5
(ii) an odd number
(iii) not an odd number
(iv) a prime number
(v) a square number
(vi) a multiple of 3 ?

Answer:
The set of counters in the bag is S = {2, 3, 4, 5, 6, 7, 8, 9, 10}. The total number of possible outcomes, \( n(S) = 9 \).
(i) Let E be the event of getting a counter with the number 5. The favourable outcome is {5}. So, \( n(E) = 1 \). The required probability \( = \frac{n(E)}{n(S)} = \frac{1}{9} \).
(ii) Let F be the event of getting a counter with an odd number. The odd numbers in the set are {3, 5, 7, 9}. So, \( n(F) = 4 \). The required probability \( = \frac{n(F)}{n(S)} = \frac{4}{9} \).
(iii) Let G be the event of getting a counter with not an odd number (which means an even number). The even numbers in the set are {2, 4, 6, 8, 10}. So, \( n(G) = 5 \). The required probability \( = \frac{n(G)}{n(S)} = \frac{5}{9} \).
(iv) Let H be the event of getting a counter with a prime number. The prime numbers in the set are {2, 3, 5, 7}. So, \( n(H) = 4 \). The required probability \( = \frac{n(H)}{n(S)} = \frac{4}{9} \).
(v) Let A be the event of getting a counter with a square number. The square numbers in the set are {4, 9} (since \( 2^2 = 4 \) and \( 3^2 = 9 \)). So, \( n(A) = 2 \). The required probability \( = \frac{n(A)}{n(S)} = \frac{2}{9} \).
(vi) Let B be the event of getting a counter that is a multiple of 3. The multiples of 3 in the set are {3, 6, 9}. So, \( n(B) = 3 \). The required probability \( = \frac{n(B)}{n(S)} = \frac{3}{9} = \frac{1}{3} \).
In simple words: When you pick one counter from a bag of nine numbered 2 to 10, the total options are 9. To find the probability of a specific type of number, count how many of those numbers are in the bag and divide by 9. For example, there are 4 odd numbers, so the chance is 4/9.

🎯 Exam Tip: Clearly list the numbers in your sample space and then filter them for each specific event (odd, prime, square, multiple) to ensure accuracy in counting favourable outcomes.

 

Question 6. A die is rolled. If the outcome is an even number, what is the probability that it is a prime number.
Answer:
This is a conditional probability problem. We are given that the outcome is an even number. So, our new sample space (S') for the even numbers on a die is {2, 4, 6}. The total number of outcomes in this new sample space, \( n(S') = 3 \). Now, let A be the event that the number is prime within this new sample space. The prime number in {2, 4, 6} is {2}. So, \( n(A) = 1 \). Therefore, the required probability \( = \frac{n(A)}{n(S')} = \frac{1}{3} \).
In simple words: If you roll a die and know the number is even (meaning it can only be 2, 4, or 6), and you want to know the chance it's a prime number, then only the number 2 in that list is prime. So, it's 1 chance out of the 3 even numbers.

🎯 Exam Tip: For conditional probability questions (starting with "if"), always reduce your sample space first to only include the outcomes that satisfy the given condition, then find the favourable outcomes within that reduced space.

 

Question 7. A bag contains 20 coloured balls. 8 are red, 6 are blue, 3 are green, 2 are white and 1 is brown. A ball is chosen at random from the bag. What is the probability that the ball chosen is :
(i) blue
(ii) not blue
(iii) brown
(iv) not brown
(v) blue or red
(vi) red or green
(vii) green or white or brown ?

Answer:
Given:
Number of red balls = 8
Number of blue balls = 6
Number of green balls = 3
Number of white balls = 2
Number of brown balls = 1
Total number of balls = \( 8 + 6 + 3 + 2 + 1 = 20 \). So, \( n(S) = 20 \).
(i) Probability of getting a blue ball:
Number of favourable outcomes (blue balls) = 6.
Required probability \( = \frac{\text{No. of favourable outcomes}}{\text{n(S)}} = \frac{6}{20} = \frac{3}{10} \).
(ii) Probability of getting not a blue ball:
Number of balls not blue = Total balls - Number of blue balls = \( 20 - 6 = 14 \).
Required probability \( = \frac{14}{20} = \frac{7}{10} \).
(iii) Probability of getting a brown ball:
Number of favourable outcomes (brown balls) = 1.
Required probability \( = \frac{1}{20} \).
(iv) Probability of getting not a brown ball:
Number of balls not brown = Total balls - Number of brown balls = \( 20 - 1 = 19 \).
Required probability \( = \frac{19}{20} \).
(v) Probability of getting a blue or red ball:
Number of favourable outcomes (blue or red) = Number of blue balls + Number of red balls = \( 6 + 8 = 14 \).
Required probability \( = \frac{14}{20} = \frac{7}{10} \).
(vi) Probability of getting a red or green ball:
Number of favourable outcomes (red or green) = Number of red balls + Number of green balls = \( 8 + 3 = 11 \).
Required probability \( = \frac{11}{20} \).
(vii) Probability of getting a green or white or brown ball:
Number of favourable outcomes (green or white or brown) = Number of green balls + Number of white balls + Number of brown balls = \( 3 + 2 + 1 = 6 \).
Required probability \( = \frac{6}{20} = \frac{3}{10} \).
In simple words: You have 20 balls in a bag. To find the chance of picking a certain color, count how many balls of that color (or colors) there are and divide by the total number of balls (20). For example, if there are 6 blue balls, the chance is 6/20.

🎯 Exam Tip: For "not" events, it's often easier to calculate 1 minus the probability of the event itself. For "or" events involving mutually exclusive outcomes, simply add the number of favourable outcomes before dividing by the total.

 

Question 8. A bag contains 20 balls. These are of three different colours : green, red and blue. A ball is chosen at random from the bag. The probability of a green ball is \( \frac{1}{4} \). The probability of a red ball is \( \frac{2}{5} \).
(i) What is the probability of a blue ball ?
(ii) How many balls are red ?
(iii) How many balls are green ?
(iv) How many balls are blue ?

Answer:
Given:
Total number of balls, \( n(S) = 20 \). Probability of getting a green ball, \( P(G) = \frac{1}{4} \). Probability of getting a red ball, \( P(R) = \frac{2}{5} \).
(i) The probability of getting a blue ball:
Since there are only three colors (green, red, and blue), the sum of their probabilities must be 1.
\( P(Blue) = 1 - P(G) - P(R) \) \( = 1 - \frac{1}{4} - \frac{2}{5} \) To subtract these fractions, find a common denominator, which is 20. \( = \frac{20}{20} - \frac{5}{20} - \frac{8}{20} \) \( = \frac{20 - 5 - 8}{20} \) \( = \frac{7}{20} \). So, the probability of choosing a blue ball is \( \frac{7}{20} \).
(ii) Number of red balls:
Let \( x \) be the number of red balls. We know that \( P(R) = \frac{\text{Number of red balls}}{\text{Total balls}} = \frac{x}{20} \). We are given \( P(R) = \frac{2}{5} \). So, \( \frac{x}{20} = \frac{2}{5} \). Multiply both sides by 20 to solve for \( x \): \( x = \frac{2}{5} \times 20 \) \( x = 2 \times 4 \) \( x = 8 \). There are 8 red balls in the bag.
(iii) Number of green balls:
Let \( y \) be the number of green balls. We know that \( P(G) = \frac{\text{Number of green balls}}{\text{Total balls}} = \frac{y}{20} \). We are given \( P(G) = \frac{1}{4} \). So, \( \frac{y}{20} = \frac{1}{4} \). Multiply both sides by 20 to solve for \( y \): \( y = \frac{1}{4} \times 20 \) \( y = 5 \). There are 5 green balls in the bag.
(iv) Number of blue balls:
We know the total number of balls is 20. Number of blue balls = Total balls - Number of red balls - Number of green balls \( = 20 - 8 - 5 \) \( = 12 - 5 \) \( = 7 \). There are 7 blue balls in the bag.
In simple words: There are 20 balls in total. If you know the chance of picking green or red balls, you can find the chance of picking a blue ball by subtracting the green and red chances from 1 (the total chance). To find the actual number of balls for each color, multiply the total balls by the probability of picking that color.

🎯 Exam Tip: Remember that the sum of probabilities of all possible outcomes for an event is always 1. Use this property to find missing probabilities. To find the number of items, multiply the total number of items by its probability.

 

Question 9. A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
Answer:
When a pair of dice is thrown, the total possible outcomes, \( n(S) = 36 \). Each outcome is a pair like (first die, second die). The sample space is: S = {(1, 1), (1,2), (1,3), (1,4), (1, 5), (1,6), (2, 1), (2,2), (2, 3), (2,4), (2, 5), (2,6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5,2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
This is a conditional probability problem: "if 5 appears on the first die." So, we need to consider only the outcomes where the first die shows a 5. Let S' be this reduced sample space: S' = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}. The number of outcomes in S' is \( n(S') = 6 \).
Now, let A be the event of getting a sum of 10 or more within S'. The outcomes in S' with a sum of 10 or more are: (5, 5) (sum = 10) (5, 6) (sum = 11) So, \( n(A) = 2 \).
The required probability \( = \frac{n(A)}{n(S')} = \frac{2}{6} = \frac{1}{3} \).
In simple words: If you roll two dice and you know the first die is a 5, then you only look at pairs like (5,1), (5,2), etc. Out of these 6 pairs, only (5,5) and (5,6) add up to 10 or more. So, the chance is 2 out of 6, which simplifies to 1 out of 3.

🎯 Exam Tip: For conditional probability, always filter the original sample space based on the "given" condition first. Then, identify the favourable outcomes from this *reduced* sample space.

 

Question 10. A match can be won, drawn or lost. One week a school is to play two matches. Draw a tree diagram to show all the possible outcomes and list them. If the outcomes are equally likely, calculate the probability that:
(i) both matches are won ;
(ii) one match is drawn ;
(iii) at least one match is drawn ;
(iv) no match is lost;
(v) both matches are not lost.

Answer:
A match can have 3 outcomes: Win (W), Draw (D), or Loss (L). For two matches, we can draw a tree diagram to show all possible outcomes:

1st Match     2nd Match     Outcome
  W --------- W             WW
  |           D             WD
  |           L             WL
  |
  D --------- W             DW
  |           D             DD
  |           L             DL
  |
  L --------- W             LW
              D             LD
              L             LL

The sample space (S) of all possible outcomes is: S = {WW, WD, WL, DW, DD, DL, LW, LD, LL}. The total number of outcomes, \( n(S) = 9 \).
(i) Let E be the event that both matches are won. Favourable outcome = {WW}. So, \( n(E) = 1 \). Required probability \( = \frac{n(E)}{n(S)} = \frac{1}{9} \).
(ii) Let F be the event that one match is drawn. Favourable outcomes = {WD, DW, DL, LD}. So, \( n(F) = 4 \). Required probability \( = \frac{n(F)}{n(S)} = \frac{4}{9} \).
(iii) Let G be the event that at least one match is drawn. This means one draw or two draws. Favourable outcomes = {WD, DW, LD, DL, DD}. So, \( n(G) = 5 \). Required probability \( = \frac{n(G)}{n(S)} = \frac{5}{9} \).
(iv) Let A be the event that no match is lost. This means only wins and draws are allowed. Favourable outcomes = {WW, WD, DW, DD}. So, \( n(A) = 4 \). Required probability \( = \frac{n(A)}{n(S)} = \frac{4}{9} \).
(v) Let B be the event that both matches are not lost. This is the same as "no match is lost" from part (iv). Favourable outcomes = {WW, WD, DW, DD}. So, \( n(B) = 4 \). Required probability \( = \frac{n(B)}{n(S)} = \frac{4}{9} \).
In simple words: When a school plays two matches, each can be a win, draw, or loss, making 9 total ways for the results to happen. To find the chance of something specific, count how many of those 9 ways match what you want, then divide by 9. For example, only 1 way means both are won, so the chance is 1/9.

🎯 Exam Tip: Drawing a tree diagram helps to visualize and list all possible outcomes systematically, which is crucial for correctly identifying the sample space and favourable events in multi-stage experiments.

 

Question 11. The ace, king, queen, jack and ten from both the spades and hearts suits are placed in two separate piles and one card is taken from each pile : Draw the sample space diagram and find the probability that:
(i) both cards will be kings ;
(ii) both of the cards could be either ace or a king ;
(iii) both cards will be a pair;
(iv) at least one card will be an ace ;
(v) neither card will be a 10;
(vi) neither card will be a king or jack;
(vii) one card will be a spade ;
(viii) both cards will be hearts.

Answer:
From spades, we have {Ace (A), King (K), Queen (Q), Jack (J), Ten (10)}. Let's call these S_A, S_K, S_Q, S_J, S_10. From hearts, we have {Ace (A), King (K), Queen (Q), Jack (J), Ten (10)}. Let's call these H_A, H_K, H_Q, H_J, H_10. One card is taken from each pile. We can represent the outcomes as (Spade card, Heart card).
The sample space diagram is:

The total sample space (S) consists of \( 5 \times 5 = 25 \) outcomes. So, \( n(S) = 25 \).
(i) Let A be the event that both drawn cards are kings. Favourable outcome = {(K, K)}. So, \( n(A) = 1 \). Required probability \( = \frac{n(A)}{n(S)} = \frac{1}{25} \).
(ii) Let B be the event that both cards could be either an ace or a king. Favourable outcomes = {(A, A), (K, K)}. So, \( n(B) = 2 \). Required probability \( = \frac{n(B)}{n(S)} = \frac{2}{25} \).
(iii) Let C be the event that both drawn cards will be a pair (same rank). Favourable outcomes = {(A, A), (K, K), (Q, Q), (J, J), (10, 10)}. So, \( n(C) = 5 \). Required probability \( = \frac{n(C)}{n(S)} = \frac{5}{25} = \frac{1}{5} \).
(iv) Let E be the event that at least one card will be an ace. This means the first card is an Ace, or the second card is an Ace, or both are Aces. Favourable outcomes = {(A, A), (A, K), (A, Q), (A, J), (A, 10), (K, A), (Q, A), (J, A), (10, A)}. So, \( n(E) = 9 \). Required probability \( = \frac{n(E)}{n(S)} = \frac{9}{25} \).
(v) Let F be the event that neither card will be a 10. This means both cards must be chosen from {A, K, Q, J}. The number of choices for the first card is 4, and for the second card is 4. So, \( n(F) = 4 \times 4 = 16 \). Required probability \( = \frac{n(F)}{n(S)} = \frac{16}{25} \).
(vi) Let G be the event that neither card will be a King or Jack. This means both cards must be chosen from {A, Q, 10}. The number of choices for the first card is 3, and for the second card is 3. So, \( n(G) = 3 \times 3 = 9 \). Required probability \( = \frac{n(G)}{n(S)} = \frac{9}{25} \).
(vii) Let H be the event that one card will be a spade. This implies one card is from the spades pile, and the other from the hearts pile. This condition is already met as we pick one card from each pile by definition. Every outcome in our sample space S has one spade and one heart. Thus, all 25 outcomes are favourable. So, \( n(H) = 25 \). Required probability \( = \frac{n(H)}{n(S)} = \frac{25}{25} = 1 \).
(viii) Let I be the event that both cards will be hearts. However, we are drawing one card from spades and one card from hearts. It's impossible to draw two heart cards from this setup. So, \( n(I) = 0 \). Required probability \( = \frac{n(I)}{n(S)} = \frac{0}{25} = 0 \).
In simple words: You pick one card from a pile of spades (A, K, Q, J, 10) and one from a pile of hearts (A, K, Q, J, 10). This gives 25 possible pairs. To find the probability of a specific outcome, count how many pairs match that outcome and divide by 25. For instance, only one pair is two Kings, so it's 1/25. It's impossible to pick two hearts, so that probability is 0.

🎯 Exam Tip: When drawing cards from separate piles, the sample space diagram (a grid or table) is an excellent way to list all \( m \times n \) possible outcomes. Pay close attention to wording like "at least one," "neither," and "from each pile" to correctly identify favourable outcomes.