\[ \begin{array} (A) \frac{37}{ 221} \quad & (B) \frac{5}{13} \quad & (C) \frac{1}{13} \quad & (D) \frac{2}{13} \end{array} \]

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- Mean of the probability distribution = $\sum (X_i \times P(X_i))$ The Expected value of X is nothing but the mean of X.

Two Cards are drawn from a deck. Let X be the number of Aces drawn.

In a deck there are 52 cards and 4 aces. We can draw 2 cards out of 52 in $^{52}C_2$ ways.

P (X = 0) = P (no aces drawn) = $\large\frac{^{48}C_2}{ ^{52}C_2} =\frac{188}{221}$

P (X = 1) = P (1 ace,1 non-ace drawn) = $\large\frac{^{48}C_1 . ^{4}C_1}{ ^{52}C_2} =\frac{32}{221}$

P (X =2) = P (two aces are drawn) = $\large\frac{^{4}C_2}{ ^{52}C_2} =\frac{1}{221}$

Therefore the mean $E(X) = 0 \times \large\frac{188}{221} $$+1\times\large\frac{32}{221}$$+2\times\large\frac{1}{221} = \frac{34}{221} = \frac{2}{13}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...