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12th Standard Maths English Medium -Probability Distributions 2 Mark Book Back Question Paper and Answer Key 2022 - 2023

Published on: 02/09/2022

QB365 provides a detailed and simple solution for every Possible Book Back Questions in Class 12 Maths Subject -Probability Distributions, English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Questions:

Generic 3

2 x 3 = 6

Suppose two coins are tossed once. If X denotes the number of tails,
(i) write down the sample space
(ii) find the inverse image of 1
(iii) the values of the random variable and number of elements in its inverse images

Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.

Generic 2

1 x 2 = 2

Compute P(X = k) for the binomial distribution, B(n, p) where
\(n=10, p=\frac{1}{5}, k=4\)

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Answers Key:

Generic 3

2 x 3 = 6

(i) The sample space S = {H,T}\(\times\){H,T}
That is S = {TT,TH,HT,HH}
(ii) Let X : S ⟶R be the number of tails
Then X (TT) = 2 (2 Tails)
X (TH ) = 1 (1 Tail)
X (HT) = 1 (1 Tail)
and X (HH) = 0 (0 Tails).
Then X is a random variable that takes on the values 0, 1 and 2.
Let X (ω) denotes the number of tails, this gives
\(\\ \\ \\ \\ \\ \\ X\left( \omega \right) =\begin{cases} \begin{matrix} 2 & if\omega =TT \end{matrix} \\ \begin{matrix} 1 & if\omega =HT,TH \end{matrix} \\ \begin{matrix} 0 & if\omega =HH \end{matrix} \end{cases}\)
The inverse images of 1 {TH, HT} . That is X^-1{1} = {TH, HT}.
(iii) Number of elements in inverse images are shown in the table.

Values of the Random Variable	0	1	2	Total
Number of elements in inverse image	1	2	1	4

Sample space s = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n(S) = 8
Let X be the random variable denoting the number of heads
X = {0, 1, 2, 3}
Values of randomn variabl 0 1 2 3 Total Number of elements in inverse image 1 3 3 1 8 Probability mass function
x 0 1 2 3 f(x) \(\cfrac { 1 }{ 8 } \) \(\cfrac { 3 }{ 8 } \) \(\cfrac { 3 }{ 8 } \) \(\cfrac { 1 }{ 8 } \) \(\therefore f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 8 } & \begin{matrix} for & x=0,3 \end{matrix} \end{matrix} \\ \begin{matrix} \frac { 3 }{ 8 } & \begin{matrix} for & x=1,2 \end{matrix} \end{matrix} \end{cases}\)

Generic 2

1 x 2 = 2

\( \therefore q =1-p=1-\frac{1}{5}=\frac{4}{5} \)
\(\mathrm{P}(\mathrm{X}=x) =n \mathrm{C}_{x} p^{x} q^{n-x}, x=0,1,2, \ldots \ldots n \)
\(\mathrm{P}(\mathrm{X}=k) =\mathrm{P}(\mathrm{X}=4) \)
\(=10 \mathrm{C}_{4}\left(\frac{1}{5}\right)^{4}\left(\frac{4}{5}\right)^{10-4}=210\left(\frac{1}{5^{4}}\right)\left(\frac{4^{6}}{5^{6}}\right)=210\left(\frac{1}{5}\right)^{4}\left(\frac{4}{5}\right)^{6} \)