Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 8 Binomial Distribution Ex 8.1 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 8 Binomial Distribution Ex 8.1
Solution & Step-by-Step Answer:
Let X = number of successes, i.e. number of odd numbers. p = probability of getting an odd number in a single throw of a die ∴ p = and q = 1 – p = 1 – = Given: n = 6 ∴ X ~ B(6, ) The p.m.f. of X is given by Hence, the probability of 5 successes is .
(ii) P(at least 5 successes) = P[X ≥ 5]
= p(5) + p(6)
Hence, the probability of at least 5 successes is .
(iii) P(at most 5 successes) = P[X ≤ 5]
= 1 – P[X > 5]
Hence, the probability of at most 5 successes is .
Solution & Step-by-Step Answer:
Let X = number of doublets. p = probability of getting a doublet when a pair of dice is thrown ∴ p = and q = 1 – p = 1 – = Given: n = 4 ∴ X ~ B(4, ) The p.m.f. of X is given by Hence, the probability of two successes is .
Solution & Step-by-Step Answer:
Let X = number of defective items. p = probability of defective item ∴ p = 5% = and q = 1 – p = 1 – = ∴ X ~ B(10, ) The p.m.f. of X is given by P(sample of 10 items will include not more than one defective item) = P[X ≤ 1] Hence, the probability that a sample of 10 items will include not more than one defective item = 29.
Solution & Step-by-Step Answer:
Let X = number of spade cards. p = probability of drawing a spade card from a pack of 52 cards. Since there are 13 spade cards in the pack of 52 cards. ∴ p = and q = 1 – p = 1 – = Given: n = 5 ∴ X ~ B(5, ) The p.m.f. of X is given by (i) P(all five cards are spade) Hence, the probability of all the five cards are spades =
(ii) P(only 3 cards are spade) = P[X = 3]
Hence, the probability of only 3 cards are spades =
(iii) P(none of cards is spade) = P[X = 0]
Hence, the probability of none of the cards is a spade =
Solution & Step-by-Step Answer:
Let X = number of fuse bulbs. p = probability of a bulb produced by a factory will fuse after 150 days of use. ∴ p = 0.05 and q = 1 – p = 1 – 0.05 = 0.95 Given: n = 5 ∴ X ~ B(5, 0.05) The p.m.f. of X is given by P(X = x) = i.e. p(x) = , x = 0, 1, 2, 3, 4, 5 (i) P(none of a bulb produced by a factory will fuse after 150 days of use) = P[X = 0] = p(0) = = 1 × 1 × (0.95)5 = (0.95)5 Hence, the probability that none of the bulbs will fuse after 150 days = (0.95)5.
(ii) P(not more than one bulb will fuse after 150 days of j use) = P[X ≤ 1]
= p(0) + p(1)
=
= 1 × 1 × (0.95)5+ 5 × (0.05) × (0.95)4
= (0.95)4[0.95 + 5(0.05)]
= (0.95)4(0.95 + 0.25)
= (0.95)4(1.20)
= (1.2) (0.95)4
Hence, the probability that not more than one bulb will fuse after 150 days = (1.2)(0.95)4.
(iii) P(more than one bulb fuse after 150 days)
= P[X > 1]
= 1 – P[X ≤ 1]
= 1 – (1.2)(0.95)4
Hence, the probability that more than one bulb fuse after 150 days = 1 – (1.2)(0.95)4.
(iv) P(at least one bulb fuse after 150 days)
= P[X ≥ 1]
= 1 – P[X = 0]
= 1 – p(0)
= 1 –
= 1 – 1 × 1 × (0.95)5
= 1 – (0.95)5
Hence, the probability that at least one bulb fuses after 150 days = 1 – (0.95)5.
Solution & Step-by-Step Answer:
Let X = number of balls marked with digit 0. p = probability of drawing a ball from 10 balls marked with the digit 0. ∴ p = and q = 1 – p = 1 – = The p.m.f. of X is given by P(none of the ball marked with digit 0) = P(X = 0) Hence, the probability that none of the bulb marked with digit 0 is
Solution & Step-by-Step Answer:
Let X = number of correct answers. p = probability that a candidate gets a correct answer from three possible answers. ∴ p = and q = 1 – p = 1 – = Given: n = 5 ∴ X ~ B(5, ) The p.m.f. of X is given by P(four or more correct answers) = P[X ≥ 4] = p(4) + p(5) Hence, the probability of getting four or more correct answers = .
Solution & Step-by-Step Answer:
Let X = number of winning prizes. p = probability of winning a prize ∴ p = and q = 1 – p = 1 – = Given: n = 50 ∴ X ~ B(50, ) The p.m.f. of X is given by i.e., p(x) = , x = 0, 1, 2,… 50 (i) P(a person wins a prize at least once) Hence, probability of winning a prize at least once = 1 –
(ii) P(a person wins exactly one prize) = P[X = 1] = p(1)
Hence, probability of winning a prize exactly once =
(iii) P(a persons wins the prize at least twice) = P[X ≥ 2]
= 1 – P[X < 2]
= 1 – [p(0) + p(1)]
Hence, the probability of winning the prize at least twice = 1 – 149.
Solution & Step-by-Step Answer:
Let X = number of working discs. p = probability that a floppy disc works ∴ p = 95% = and q = 1 – p = 1 – = Given: n = 3 ∴ X ~ B(3, ) The p.m.f. of X is given by (i) P(none of the floppy discs work) = P(X = 0) Hence, the probability that none of the floppy disc will work = .
(ii) P(exactly one floppy disc works) = P(X = 1)
Hence, the probability that exactly one floppy disc works = 3
(iii) P(exactly two floppy discs work) = P(X = 2)
Hence, the probability that exactly 2 floppy discs work = 3
(iv) P(all 3 floppy discs work) = P(X = 3)
Hence, the probability that all 3 floppy discs work = .
Solution & Step-by-Step Answer:
Let X = number of sixes. p = probability that a die shows six in a single throw ∴ p = and q = 1 – p = 1 – = Given: n = 6 ∴ X ~ B(6, ) The p.m.f. of X is given by Hence, probability of throwing at most 2 sixes = .
Solution & Step-by-Step Answer:
Let X = number of defective articles. p = probability of defective articles. ∴ p = 10% = and q = 1 – p = 1 – = Given: n = 12 ∴ X ~ B(12, ) The p.m.f. of X is given by Hence, the probability of getting 9 defective articles =
Solution & Step-by-Step Answer:
(i) Given: n = 10 and p = 0.4 ∴ q = 1 – p = 1 – 0.4 = 0.6 ∴ E(X) = np = 10(0.4) = 4 Var(X) = npq = 10(0.4)(0.6) = 2.4 Hence, E(X) = 4, Var(X) = 2.4.
(ii) Given: p = 0.6, E (X) = 6
E(X) = np
6 = n(0.6)
n = = 10
Now, q = 1 – p = 1 – 0.6 = 0.4
∴ Var(X) = npq = 10(0.6)(0.4) = 2.4
Hence, n = 10 and Var(X) = 2.4.
(iii) Given: n = 25, E(X) = 10
E(X) = np
10 = 25p
p =
∴ q = 1 – p = 1 – =
Var(X) = npq = = 6
∴ SD(X) = √Var(X) = √6
Hence, p = and S.D.(X) = √6.
(iv) Given: n = 10, E(X) = 8
E(X) = np
8 = 10p
p =
q = 1 – p = 1 – =
Var(X) = npq =
Hence, Var(X) = .