Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Two natural numbers $r$ , $s$ are drawn one at a time, without replacement from the set $S = {1, 2, 3, . . ., n}$ . Find $P [r \leq p \leq s ∣ p \in S]$

10 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : S = {1, 2, 3, \dots, n} \\ ∴ P (r \leq p / s \leq p) = \frac{P (P \cap S)}{P (S)} = \frac{p - 1}{n} \times \frac{n}{n - 1} = \frac{p - 1}{n - 1} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{p - 1}{n - 1} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Suppose that 6% of the people with blood group O are left-handed and 10% of those with other blood groups are left-handed. 30% of the people have blood group O. If a left-handed person is selected at random, what is the probability that he/she will have blood group O?

10 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} = T h e e v e n t t h a t a p e r s o n s e l e c t e d i s o f b l o o d g r o u p O \\ E_{2} = T h e e v e n t t h a t a p e r s o n s e l e c t e d i s o f o t h e r g r o u p \\ a n d H = T h e e v e n t t h a t s e l e c t e d p e r s o n i s l e f t h a n d e d . \\ P (E_{1}) = 0.30, P (E_{2}) = 0.70, P (H / E_{1}) = 0.06, P (H / E_{2}) = 0.10 \\ ∴ Using B a y e' s T h e o r m, w e g e t \\ P (E_{1} / H) = \frac{P (E_{1}) . P (H / E_{1})}{P (E_{1}) . P (H / E_{1}) + P (E_{2}) . P (H / E_{2})} \\ = \frac{0.30 \times 0.06}{0.30 \times 0.06 + 0.70 \times 0.10} = \frac{0.018}{0.018 + 0.070} = \frac{0.018}{0.088} = \frac{9}{44} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{9}{44} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} = E v e n t t h a t t h e c o i n i s f a i r \\ E_{2} = E v e n t t h a t t h e c o i n i s 2 - h e a d e d \\ a n d H = E v e n t t h a t t h e t o s s e d c o i n g e t s h e a d . \\ P (E_{1}) = \frac{1}{2}, P (E_{2}) = \frac{1}{2}, P (H / E_{1}) = \frac{1}{2}, P (H / E_{2}) = 1 \\ ∴ Using B a y e' s T h e o r m, w e g e t \\ P (E_{1} / H) = \frac{P (E_{1}) . P (H / E_{1})}{P (E_{1}) . P (H / E_{1}) + P (E_{2}) . P (H / E_{2})} \\ = \frac{\frac{1}{2} . \frac{1}{2}}{\frac{1}{2} . \frac{1}{2} + \frac{1}{2} . 1} = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{1}{3} . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A factory produces bulbs. The probability that any one bulb is defective is $\frac{1}{50}$ and they are packed in boxes of 10. From a single box, find the probability that:

i. one of the bulbs is defective

ii. exactly two bulbs are defective

iii. more than 8 bulbs work properly

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t X b e t h e r a n d o m d e n o t i n g a b u l b t o b e d e f e c t i v e . \\ H e r e, n = 10, p = \frac{1}{50}, q = 1 - \frac{1}{50} = \frac{49}{50} \\ W e k n o w t h a t P (X = r) = {}_{n}{C_{r}} p^{r} q^{n - r} \\ (i) N o n e o f t h e b u l b s i s d e f e c t i v e, i . e ., r = 0 \\ P (x = 0) = {}_{10}{C_{0}} {(\frac{1}{50})}^{0} {(\frac{49}{50})}^{10 - 0} = {(\frac{49}{50})}^{10} \\ (i i) E x a c t l y t w o b u l b s a r e d e f e c t i v e, i . e ., r = 2 \\ ∴ P (x = 2) = {}_{10}{C_{2}} {(\frac{1}{50})}^{2} {(\frac{49}{50})}^{10 - 2} = 45 . \frac{{(49)}^{8}}{{(50)}^{10}} = 45 \times {(\frac{1}{50})}^{10} \times {(49)}^{8} \\ (i i i) M o r e t h a n 8 b u l b s w o r k p r o p e r l y \\ W e c a n s a y t h a t l e s s t h a n 2 b u l b s a r e d e f e c t i v e \\ P (X < 2) = P (x = 0) + P (x = 1) \\ = {}_{10}{C_{0}} {(\frac{1}{50})}^{0} {(\frac{49}{50})}^{10 - 0} + {}_{10}{C_{1}} {(\frac{1}{50})}^{1} {(\frac{49}{50})}^{9} = {(\frac{49}{50})}^{10} + \frac{1}{5} {(\frac{49}{50})}^{9} \\ = {(\frac{49}{50})}^{9} (\frac{49}{50} + \frac{1}{5}) = {(\frac{49}{50})}^{9} (\frac{59}{50}) = \frac{59 {(49)}^{9}}{{(50)}^{10}} \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Two probability distributions of the discrete random variable X and Y are given below:

X

0

1

2

3

P(X)

Y

0

1

2

3

P(Y)

Prove that $E (Y^{2}) = 2 E (X)$ .

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol: $\begin{array}{l} \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

Two biased dice are thrown together. For the first die $P (6) = \frac{1}{2}$ , the other scores being equally likely, while for the second die, $P (1) = \frac{2}{5}$ and the other scores are equally likely. Find the probability distribution of ‘the number of ones seen’.

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT

Sol:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, w e h a v e X = 0, 1, 2, 3 [? d i e i s t h r o w n 3 t i m e s] \\ a n d p = \frac{1}{6}, q = 1 - \frac{1}{6} = \frac{5}{6} \\ ∴ P (X = 0) = P (n o t 2) . P (n o t 2) . P (n o t 2) = \frac{5}{6} . \frac{5}{6} . \frac{5}{6} = \frac{125}{216} \\ P (X = 1) = P (2) . P (2) . P (n o t 2) + P (2) . P (n o t 2) . P (2) + P (n o t 2) . P (2) . P (2) \\ = \frac{1}{6} . \frac{1}{6} . \frac{5}{6} + \frac{1}{6} . \frac{5}{6} . \frac{1}{6} + \frac{5}{6} . \frac{1}{6} . \frac{1}{6} = \frac{5}{216} + \frac{5}{216} + \frac{5}{216} = \frac{15}{216} \\ P (X = 3) = P (2) . P (2) . P (2) = \frac{1}{6} . \frac{1}{6} . \frac{1}{6} = \frac{1}{216} \\ N o w, E (X) = \sum_{i = 1}^{n} p_{i} x_{i} \\ = 0 \times \frac{125}{216} + 1 \times \frac{75}{216} + 2 \times \frac{15}{216} + 3 \times \frac{1}{216} \\ = 0 + \frac{75}{216} + \frac{30}{216} + \frac{3}{216} = \frac{75 + 30 + 3}{216} = \frac{108}{216} = \frac{1}{2} \\ H e n c e, t h e r e q u i r e d i s \frac{1}{2} . \end{array}$