Students often turn to S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Ex 22(e) to clarify doubts and improve problem-solving skills.
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 ?
Solution:
(i) When a die is thrown then total no. of outcomes = n (S) = 6 where S = {1, 2,3, 4, 5, 6}
∴ P (getting 2) = \(\frac { 1 }{ 6 }\)
(ii) P (getting 5) = \(\frac { 1 }{ 6 }\)
(iii) P (getting 2 or 5) = P (2) + P (5) = \(\frac { 1 }{ 6 }\) + \(\frac { 1 }{ 6 }\) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)
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 ?
Solution:
Total no. of cards = n (S) = 52
(i) since there are 13 heart cards ∴ P (heart) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(ii) p (a club) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(iii) P (a heart or a club) = P (heart) + P (club) = \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 2 }\)
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
Solution:
Here n (S) = 52
(i) Since there are four 3’s (one each from heart, diamond, club and spade)
∴ required probability of getting 3 = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(ii) P (getting 7) = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(iii) P (3 or 7) = P (3) + P (7) = \(\frac { 1 }{ 13 }\) = \(\frac { 1 }{ 13 }\) = \(\frac { 2 }{ 13 }\)
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 ?
Solution:
Total no. of letters in word PROBABILITY = 11
(i) P (getting a letter B) = \(\frac { 2 }{ 11 }\) [since there are 2B’s]
(ii) since there are 4 vowels O, A, I, I
∴ P (getting a vowel) = \(\frac { 4 }{ 11 }\)
(iii) P (B or a vowel) = P (B) + P (a vowel) = \(\frac { 2 }{ 11 }\) + \(\frac { 4 }{ 11 }\) = \(\frac { 6 }{ 11 }\)
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 ?
Solution:
Given no. of white balls = 7, no. of green balls = 9 and no. of yellow balls = 10
∴ Total no. of balls = 7 + 9 + 10 = 26
(i) P(white ball) = \(\frac { 7 }{ 26 }\)
(ii) P(green ball) = \(\frac { 9 }{ 26 }\)
(iii) P(green or white) = P(green) + P(white) = \(\frac { 9 }{ 26 }\) + \(\frac { 7 }{ 26 }\) = \(\frac { 16 }{ 26 }\) = \(\frac { 8 }{ 13 }\)
(iv) P(yellow ball) = \(\frac { 10 }{ 26 }\) ∴ P(getting not yellow ball) = 1 – \(\frac { 10 }{ 26 }\) = \(\frac { 16 }{ 26 }\) = \(\frac { 8 }{ 13 }\)
Thus, P (neither yellow nor green) = P (white ball) = \(\frac { 7 }{ 26 }\)
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 ?
Solution:
(i) Given P (calculator is in his pocket) = 0.20
P (calculator is in his bag) = 0.58
∴ required probability = P (calculator in his pocket) + P (calculator in his bag) = 0.20 + 0.58 = 0.78
(iii) required probability that calculator is in his locker = 1 – P (calculator is in his pocket) – P (calculator is in his bag) = 1 – 0.20 – 0.58 = 0.22
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 following ?
(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.
Solution:
Given P (red) = 0.5 ; P (1) = 0.4
P (green) = 0.25 ; P (2) = 0.35
P (blue) = 0.25 ; P (3) = 0.25
(i) P (red or green) = P (red) + P (green) = 0.5 + 0.25 = 0.75
(ii) P (2 or 3) = P (2) + P (3) = 0.35 + 0.25 = 0.6
(iii) P (3 or green) = P (3) + P (green) = 0.25 + 0.25 = 0.50
(iv) P (2 or green) = P (2) + P (green) = 0.35 + 0.25 = 0.60
(v) Here 1 and red are not mutually exclusive event
∴ P (1 and red) ≠ 0 [∵ 1 ∩ red ≠ Φ]
∴ P (1 or red) ≠ P(1) + P(red) = 0.9