Match List – I with List – II.
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LIST I |
LIST II |
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A. $x_1+x_2=2$ and $P\left(x_1\right)= P\left(x_2\right)=\frac{1}{2}$, then $E(x)$ |
I. $\frac{4}{3}$ |
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B. The mean of a Binomial distribution $B\left(4, \frac{1}{3}\right)$ |
II. $\frac{5}{12}$ |
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C. Variance of probability distribution of number of sixes in 3 throws of a die |
III. $\frac{1}{2}$ |
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D. Mean of probability distribution of number of sixes in 3 throws of a die |
IV. 1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A - IV, B - I, C - II, D - III
$\text{(A)}\; x_1+x_2=2,\; P=\frac{1}{2}$
$E(X)=\frac{x_1+x_2}{2}=1 \Rightarrow \text{(IV)}$
$\text{(B)}\; B(4,\frac{1}{3})$
$\text{Mean}=np=4\cdot\frac{1}{3}=\frac{4}{3} \Rightarrow \text{(I)}$
$\text{(C)}\; \text{Variance of Binomial } (n=3,p=\frac{1}{6})$
$npq=3\cdot\frac{1}{6}\cdot\frac{5}{6}=\frac{15}{36}=\frac{5}{12} \Rightarrow \text{(II)}$
$\text{(D)}\; \text{Mean}=np=3\cdot\frac{1}{6}=\frac{1}{2} \Rightarrow \text{(III)}$
A–IV,\; B–I,\; C–II,\; D–III