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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?

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alok kumar singh

Contributor-Level 10

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

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

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

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alok kumar singh

Contributor-Level 10

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

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'.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT

Sol:

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New answer posted

a year ago

Class 12th Probability

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New question posted

a year ago

Class 12th Probability

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

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New answer posted

a year ago

Class 12th Probability

A biased die is such that $P (4) = \frac{1}{10}$ and other scores being equally likely. The die is tossed twice. If X is the 'number of fours seen', find the variance of the random variable X.

0 Follower 8 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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New answer posted

a year ago

Class 12th Probability

For the following probability distribution, determine the standard deviation of the random variable X.

X	2	3	4
P(X)	0.2	0.5	0.3

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alok kumar singh

Contributor-Level 10

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

Class 12th Probability

The probability distribution of a random variable X is given below:

X	0	1	2	3
P(X)	k

i. Determine the value of k.

ii. Determine P $(x \leq 2)$ and P $(x > 2)$ .

iii. Find $P(x \leq 2) +P (x > 2)$ .

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} (i) W e k n o w t h a t P (0) + P (1) + P (2) + P (3) = 1 \\ \Rightarrow k + \frac{k}{2} + \frac{k}{4} + \frac{k}{8} = 1 \\ \Rightarrow \frac{8 k + 4 k + 2 k + k}{8} = 1 \Rightarrow 15 k = 8 \\ ∴ k = \frac{8}{15} \\ H e r e, n = 8, p = \frac{1}{10}, q = 1 - \frac{1}{10} = \frac{9}{10} \\ (i i) P (X \leq 2) = P (X = 0) + P (X = 1) + P (X = 2) \\ = k + \frac{k}{2} + \frac{k}{4} = \frac{7 k}{4} = \frac{7}{4} * \frac{8}{15} = \frac{14}{15} \\ a n d P (X > 2) = P (X = 3) = \frac{k}{8} = \frac{1}{8} * \frac{8}{15} = \frac{1}{15} \\ (i i i) P (X \leq 2) + P (X > 2) = \frac{14}{15} + \frac{1}{15} = \frac{15}{15} = 1 . \end{array}$

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New answer posted

a year ago

Class 12th Probability

Consider the probability distribution of a random variable X:

X	0	1	2	3	4
P(X)	0.1	0.25	0.3	0.2	0.15

Calculate:(i) $V (\frac{X}{2})$ (ii) Variance of X.

0 Follower 11 Views

alok kumar singh

Contributor-Level 10

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

Sol:

0 0

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