Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 8 Probability Distributions Ex 8.3 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3
Question 1
Maharashtra Board Solution
A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of (i) 2 successes (ii) at least 3 successes (iii) at most 2 successes.
Solution & Step-by-Step Answer:
X: Getting an odd no. p: Probability of getting an odd no. A die is thrown 4 times ∴ n = 4 ∵ p = ∴ q = 1 – p = 1 – = ∵ X ~ B(3, ) ∴ p(x) = (i) P(Two Successes) (ii) P(Atleast 3 Successes) (iii) P(Atmost 2 Successes)
Question 2
Maharashtra Board Solution
A pair of dice is thrown 3 times. If getting a doublet is considered a success, find the probability of two successes.
Solution & Step-by-Step Answer:
n: No. of times die is thrown = 3 X: No. of doublets p: Probability of getting doublets Getting a doublet means, same no. is obtained on 2 throws of a die There are 36 outcomes No. of doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Question 3
Maharashtra Board Solution
There are 10% defective items in a large bulk of items. What is the probability that a sample of 4 items will include not more than one defective item?
Solution & Step-by-Step Answer:
n: No of sample items = 4 X: No of defective items p: Probability of getting defective items ∴ p = 0.1 ∴ q = 1 – p = 1 – 0.1 = 0.9 X ~ B(4, 0.1) ∴ p(x) = P(Not include more than 1 defective) P(X ≤ 1) = p(0) + p(1) = 4C0 (0.1)0 (0.9)4 + 4C1 (0.1)1 (0.9)4-1 = 1 × 1 × (0.9)4 + 4 × 0.1 × (0.9)3 = (0.9)3 [0.9 + 0.4] = (0.9)3 × 1.3 = 0.977
Question 4
Maharashtra Board Solution
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability that (i) all the five cards are spades, (ii) only 3 cards are spades, (iii) none is a spade.
Solution & Step-by-Step Answer:
X: No. of spade cards Number of cards drawn ∴ n = 5 p: Probability of getting spade card (i) P(All five cards are spades) (ii) P(Only 3 cards are spades) (iii) P(None is a spade)
Question 5
Maharashtra Board Solution
The probability that a bulb produced by a factory will use fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5) that fuse after 200 days of use. Find the probability of (i) X = 0, (ii) X ≤ 1, (iii) X > 1, (iv) X ≥ 1.
Solution & Step-by-Step Answer:
X : No. of bulbs fuse after 200 days of use p : Probability of getting fuse bulbs No. of bulbs in a sample ∴ n = 5 ∴ p = 0.2 ∴ q = 1 – p = 1 – 0.2 = 0.8 ∵ X ~ B(5, 0.2) ∴ p(x) = (i) P(X = 0) = 5C0 (0.2)0 (0.8)5-0 = 1 × 1 × (0.8)5 = (0.8)5
(ii) P(X ≤ 1) = p(0) + p(1)
=5C0(0.2)0(0.8)5-0+5C1(0.2)1(0.8)5-1
= 1 × 1 × (0.8)5+ 5 × 0.2 × (0.8)4
= (0.8)4[0.8 + 1]
= 1.8 × (0.8)4
(iii) P(X > 1) = 1 – [p(0) + p(1)]
= 1 – 1.8 × (0.8)4
(iv) P(X ≥ 1) = 1 – p(0)
= 1 – (0.8)5
Question 6
Maharashtra Board Solution
10 balls are marked with digits 0 to 9. If four balls are selected with replacement. What is the probability that none is marked 0?
Solution & Step-by-Step Answer:
X : No. of balls drawn marked with the digit 0 n : No. of balls drawn ∴ n = 4 p : Probability of balls marked with 0. ∴ p = ∴ q = 1 – p = 1 – = p(x) = P(None of the ball is marked with digit 0)
Question 7
Maharashtra Board Solution
In a multiple-choice test with three possible answers for each of the five questions, what is the probability of a candidate getting four or more correct answers by random choice?
Solution & Step-by-Step Answer:
n: No. of Questions ∴ n = 5 X: No. of correct answers by guessing p: Probability of getting correct answers
Question 8
Maharashtra Board Solution
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution & Step-by-Step Answer:
X : No. of sixes in 6 throws n : No. of times dice thrown ∴ n = 6 p : Probability of getting six ∴ p = ∴ q = 1 – p = 1 – = ∵ X ~ B(6, ) ∴ p(x) = P(At most 2 sixes) P(X ≤ 2) = p(0) + p(1) + p(2)
Question 9
Maharashtra Board Solution
Given that X ~ B(n, p), (i) if n = 10 and p = 0.4, find E(X) and Var(X). (ii) if p = 0.6 and E(X) = 6, find n and Var(X). (iii) if n = 25, E(X) = 10, find p and Var(X). (iv) if n = 10, E(X) = 8, find Var(X).
Solution & Step-by-Step Answer:
∵ X ~ B (n, p), E(X) = np, V(X) = npq, q = 1 – p (i) E(X) = np = 10 × 0.4 = 4 ∵ q = 1 – p = 1 – 0.4 = 0.6 V(X) = npq = 10 × 0.4 × 0.6 = 2.4
(ii) ∵ p = 0.6
∴ q = 1 – p = 1 – 0.6 = 0.4
E(X) = np
∴ 6 = n × 0.6
∴ n = 10
∴ V(X) = npq = 10 × 0.6 × 0.4 = 2.4
(iii) E(X) = np
∴ 10 = 25 × p
∴ p = 0.4
∴ q = 1, p = 1 – 0.4 = 0.6
∴ S.D.(X) = √V(X)
=
=
= √6
= 2.4494
(iv) ∵ E(X) = np
∴ 8 = 10p
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
∵ V(X) = npq = 10 × 0.8 × 0.2 = 1.6