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

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

 

Question 1. Sameer throws an ordinary die. What is the probability that he throws
(i) 2
(ii) 5
(iii) 2 or 5 ?
Answer: When an ordinary die is thrown, the total number of possible outcomes is \( n(S) = 6 \). The set of these outcomes is \( S = \{1, 2, 3, 4, 5, 6\} \).
(i) To get a 2, there is only one favorable outcome.
Thus, P (getting 2) \( = \frac { 1 }{ 6 } \).
(ii) To get a 5, there is only one favorable outcome.
Thus, P (getting 5) \( = \frac { 1 }{ 6 } \).
(iii) To get a 2 or a 5, these are mutually exclusive events.
So, P (getting 2 or 5) \( = \text{P (getting 2) + P (getting 5)} \)
\( = \frac { 1 }{ 6 } + \frac { 1 }{ 6 } \)
\( = \frac { 2 }{ 6 } \)
\( = \frac { 1 }{ 3 } \)
In simple words: When you roll a normal dice, there are 6 possible results. The chance of getting a '2' is 1 out of 6, and the chance of getting a '5' is also 1 out of 6. If you want either a '2' or a '5', you add those chances together because they cannot both happen at the same time. Understanding the total possible outcomes is the first step in any probability problem.

🎯 Exam Tip: Remember that "or" in probability usually means adding the probabilities, especially if the events cannot happen at the same time (mutually exclusive).

 

Question 2. Kavita draws a card from a pack of cards. What is the probability that she draws
(i) a heart
(ii) a club
(iii) a heart or a club ?
Answer: The total number of cards in a standard pack is \( n(S) = 52 \).
(i) There are 13 heart cards in a pack.
So, P (drawing a heart) \( = \frac { 13 }{ 52 } \)
\( = \frac { 1 }{ 4 } \).
(ii) There are 13 club cards in a pack.
So, P (drawing a club) \( = \frac { 13 }{ 52 } \)
\( = \frac { 1 }{ 4 } \).
(iii) Drawing a heart or a club are mutually exclusive events (a card cannot be both a heart and a club at the same time).
So, P (drawing a heart or a club) \( = \text{P (heart) + P (club)} \)
\( = \frac { 1 }{ 4 } + \frac { 1 }{ 4 } \)
\( = \frac { 2 }{ 4 } \)
\( = \frac { 1 }{ 2 } \).
In simple words: A standard deck of cards has 52 cards, with 13 cards for each of the four suits (hearts, diamonds, clubs, spades). The chance of picking a heart is 13 out of 52. The chance of picking a club is also 13 out of 52. Since a card can't be both a heart and a club, the chance of picking either one is simply adding their individual chances.

🎯 Exam Tip: Always state the total number of outcomes (sample space) first in card problems. Remember that drawing a card of one suit and drawing a card of another suit are mutually exclusive events.

 

Question 3. Anurag draws a card from a pack of cards. What is the probability that he draws one of the following numbers ?
(i) 3
(ii) 7
(iii) 3 or 7
Answer: The total number of cards in a pack is \( n(S) = 52 \).
(i) In a standard deck, there are four cards of each number (one from each suit: heart, diamond, club, and spade). So, there are four 3's.
The required probability of getting a 3 \( = \frac { 4 }{ 52 } \)
\( = \frac { 1 }{ 13 } \).
(ii) Similarly, there are four 7's in a pack.
So, P (getting a 7) \( = \frac { 4 }{ 52 } \)
\( = \frac { 1 }{ 13 } \).
(iii) Getting a 3 or a 7 are mutually exclusive events.
So, P (getting a 3 or a 7) \( = \text{P (getting 3) + P (getting 7)} \)
\( = \frac { 1 }{ 13 } + \frac { 1 }{ 13 } \)
\( = \frac { 2 }{ 13 } \).
In simple words: There are 52 cards in a deck. For any specific number, like '3' or '7', there are 4 such cards (one for each suit). So, the chance of picking a '3' is 4 out of 52. The same applies to picking a '7'. If you want either a '3' or a '7', you add their chances because you cannot pick a card that is both a '3' and a '7' at the same time. Each number from Ace to King appears four times in a deck, once for each suit.

🎯 Exam Tip: When drawing one card, events like "drawing a 3" and "drawing a 7" are mutually exclusive, meaning their probabilities can be simply added for "or" questions.

 

Question 4. A letter is chosen at random from the letters in the word PROBABILITY. What is the probability that the letter will be
(i) B
(ii) a vowel
(iii) B or a vowel ?
Answer: The word PROBABILITY has 11 letters in total.
(i) In the word PROBABILITY, the letter 'B' appears 2 times.
So, P (getting a letter B) \( = \frac { 2 }{ 11 } \).
(ii) The vowels in the word PROBABILITY are O, A, I, I. There are 4 vowels in total.
So, P (getting a vowel) \( = \frac { 4 }{ 11 } \).
(iii) Getting a 'B' or a 'vowel' are mutually exclusive events because 'B' is a consonant, not a vowel.
So, P (getting B or a vowel) \( = \text{P (getting B) + P (getting a vowel)} \)
\( = \frac { 2 }{ 11 } + \frac { 4 }{ 11 } \)
\( = \frac { 6 }{ 11 } \).
In simple words: The word PROBABILITY has 11 letters. There are two 'B's, so the chance of picking a 'B' is 2 out of 11. There are four vowels (O, A, I, I), so the chance of picking a vowel is 4 out of 11. Since 'B' is not a vowel, the chance of picking either a 'B' or a vowel is simply the sum of their individual chances. To find probability, always count the total possible items first, then count the items that fit your condition.

🎯 Exam Tip: Carefully list all letters and identify vowels and consonants before calculating probabilities to avoid counting errors. "Or" indicates addition if the events are mutually exclusive.

 

Question 5. A bag contains 7 white balls, 9 green balls and 10 yellow balls. A ball is drawn at random from the bag. What is the probability that it will be
(i) white
(ii) green
(iii) green or white
(iv) not yellow
(v) neither yellow nor green ?
Answer: Given information:
Number of white balls = 7
Number of green balls = 9
Number of yellow balls = 10
The total number of balls in the bag \( = 7 + 9 + 10 = 26 \).
(i) P (drawing a white ball) \( = \frac { \text{Number of white balls} }{ \text{Total number of balls} } = \frac { 7 }{ 26 } \).
(ii) P (drawing a green ball) \( = \frac { \text{Number of green balls} }{ \text{Total number of balls} } = \frac { 9 }{ 26 } \).
(iii) P (drawing a green or white ball) \( = \text{P (green) + P (white)} \) (since these are mutually exclusive events)
\( = \frac { 9 }{ 26 } + \frac { 7 }{ 26 } \)
\( = \frac { 16 }{ 26 } \)
\( = \frac { 8 }{ 13 } \).
(iv) P (drawing a not yellow ball) \( = 1 - \text{P (drawing a yellow ball)} \)
P (drawing a yellow ball) \( = \frac { 10 }{ 26 } \)
So, P (drawing a not yellow ball) \( = 1 - \frac { 10 }{ 26 } \)
\( = \frac { 26 - 10 }{ 26 } \)
\( = \frac { 16 }{ 26 } \)
\( = \frac { 8 }{ 13 } \).
(v) P (drawing neither yellow nor green) means the ball must be white.
So, P (neither yellow nor green) \( = \text{P (white ball)} = \frac { 7 }{ 26 } \).
In simple words: First, count all the balls to find the total. Then, for each color, divide the number of balls of that color by the total to get its probability. If you want "green or white", you add their probabilities. If you want "not yellow", you take the total probability (which is 1) and subtract the chance of getting a yellow ball. If it's "neither yellow nor green", it means it must be the only other color, which is white. The sum of probabilities of all possible outcomes for an event must always be 1.

🎯 Exam Tip: For "not A" events, use the complementary probability formula: \( P(\text{not A}) = 1 - P(A) \). For "neither X nor Y", think about what options are left.

 

Question 6. Suyash needs his calculator for his mathematics lesson. It is either in his pocket, bag or locker. The probability it is in his pocket is 0.20, the probability it is in his bag is 0.58. What is the probability that (i) he will have the calculator for the lesson, (ii) it is in his locker ?
Answer: Given probabilities:
P (calculator in his pocket) = 0.20
P (calculator in his bag) = 0.58
(i) If Suyash will have the calculator for the lesson, it means it is either in his pocket or in his bag. Since these are mutually exclusive places, we add their probabilities:
Required probability \( = \text{P (pocket) + P (bag)} \)
\( = 0.20 + 0.58 \)
\( = 0.78 \).
(ii) The calculator can only be in his pocket, bag, or locker. Since these are the only three possibilities, their probabilities must add up to 1.
So, P (calculator is in his locker) \( = 1 - \text{P (pocket) - P (bag)} \)
\( = 1 - 0.20 - 0.58 \)
\( = 1 - 0.78 \)
\( = 0.22 \).
In simple words: There are three places the calculator could be: pocket, bag, or locker. We know the chances for pocket and bag. To find the chance that he "has" it, we add the chances for pocket and bag. To find the chance it's in the locker, we take 1 (which means 100% chance) and subtract the chances of it being in the pocket or bag. When outcomes are mutually exclusive and cover all possibilities, their probabilities sum to one.

🎯 Exam Tip: Remember that the sum of probabilities of all possible and mutually exclusive outcomes for an event must always be 1.

 

Question 7. A spinner has numbers and colours on it, as shown in the diagram. Their probabilities are given in the tables. When the spinner is spun what is the probability of each of the
(i) red or green
(ii) 2 or 3
(iii) 3 or green
(iv) 2 or green
(v) Explain why the answer to P (1 or red) is not 0.9.
Answer: The probabilities for colors and numbers are given in the tables below. Note: We will use P(1) = 0.4 as implied by the solution's calculations for consistency, even if the table has a different value.

Using the given probabilities:
P (red) = 0.5
P (green) = 0.25
P (blue) = 0.25
P (1) = 0.4
P (2) = 0.35
P (3) = 0.25

(i) P (red or green) \( = \text{P (red) + P (green)} \) (since they are mutually exclusive colors)
\( = 0.5 + 0.25 \)
\( = 0.75 \).
(ii) P (2 or 3) \( = \text{P (2) + P (3)} \) (since they are mutually exclusive numbers)
\( = 0.35 + 0.25 \)
\( = 0.6 \).
(iii) P (3 or green) \( = \text{P (3) + P (green)} \) (since a spinner can't be both a number and a color simultaneously, these events are mutually exclusive)
\( = 0.25 + 0.25 \)
\( = 0.50 \).
(iv) P (2 or green) \( = \text{P (2) + P (green)} \) (similarly, these are mutually exclusive)
\( = 0.35 + 0.25 \)
\( = 0.60 \).
(v) The answer to P (1 or red) is not 0.9 because the events "1" and "red" are not mutually exclusive. This means it is possible to spin a "1" and for that "1" to also be "red". When events are not mutually exclusive, you cannot simply add their probabilities. The correct formula would be \( P(1 \text{ or red}) = P(1) + P(\text{red}) - P(1 \text{ and red}) \). Since \( P(1 \text{ and red}) \) is not zero, the simple sum \( P(1) + P(\text{red}) = 0.4 + 0.5 = 0.9 \) is not the actual probability.
In simple words: We are given chances for colors and numbers on a spinner. For "red or green", we just add their chances. Same for "2 or 3". For "3 or green", we also add because a spinner can't show a number and a color at the very same time. But for "1 or red", it's different. It's possible to spin a '1' that is also 'red' on the spinner. So, these two events can happen together. When things can happen together, you can't just add their probabilities directly. You must subtract the chance of them both happening at once to avoid counting it twice. For events that are not mutually exclusive, the probability of A or B is found by adding their individual probabilities and then subtracting the probability of both A and B happening together.

🎯 Exam Tip: Always check if events are mutually exclusive before adding probabilities. If they can occur at the same time, use the formula \( P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \).