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Problems on Expectation|Statistics|UGC|Class 12

Q. A bag contains 5 white balls and 3 black balls. 3 balls are drawn at randomly without replacement. If X is random variable, which takes value 1 if at least 2 white balls are drawn, and value 0, otherwise, find E(X).

Ans: \small X_1=1 if the number of white balls are at least 2. \small P(X_1 \geq 2)=\frac{^5C_2.^3C_1}{^8C_3}+\frac{^5C_3.^3C_0}{^8C_3}=\frac{40}{56}

When 

\small X_1    : 1

\small P(X_1):\frac {40}{56}

\small X_2=0 if the number of white balls are must be less than 2. \small P(X_2 < 2)=\frac{^5C_1.^3C_2}{^8C_3}+\frac{^5C_0.^3C_3}{^8C_3}=\frac{16}{56}

When 

\small X_2     : 0

\small P(X_2):\frac {16}{56}

\small E(X)=\sum_{i=1}^{2} X_iP(X_i)=1.\frac{40}{56}+0.\frac{16}{56}=\frac{5}{7}


Q. If X and Y are both negative random variables, mutually independent. E(XY)=6 and |E(X)|=2, Find E(Y).

Ans: As we know X,Y are both negative and mutually independent so \small E(XY)=E(X)E(Y)

\small E(X)E(Y)=6 and \small |E(X)|=2 and we know X,Y are both negative random variable so, 

\small E(X)=-2 and \small E(Y)=\frac {E(XY)}{E(X)}=-\frac{6}{2}=-3 


Q.Show that \small [E(X^2)]^{\frac{1}{2}} \geq E(X).

Ans: We know that \small V(X) \geq 0

or,\small E(X^2)\geq [E(X)]^2

or,\small [E(X^2)]^{\frac{1}{2}} \geq E(X)


Q.Show that mean deviation can not exceed standard deviation.

Ans:

We know that \small V(X) \geq 0

or,\small E(X^2)\geq [E(X)]^2

let \small X=Y-\bar Y

or, \small E(Y-\bar Y)^2\geq [E(Y-\bar Y)]^2

or,\small \sqrt {E(Y-\bar Y)^2} \geq |E(Y-\bar Y)| \geq E|(Y-\bar Y)|

or, Standard deviation \small \geq Mean deviation


Q. If a person gets Rs. 2x+5, where x denotes the number appearing when a balanced die is rolled once, how much money can he except in the long run per game.

Ans: 

The outcomes from a throwing of a balanced die is: {1,2,3,4,5,6}


\small x
[value of the outcome]
\small y=2x+5
[Money]
\small P_X(x)
17\frac{1}{6}
29\frac{1}{6}
311\frac{1}{6}
413\frac{1}{6}
515\frac{1}{6}
617\frac{1}{6}

Excepted money he can earn: \small 7\times \frac{1}{6}+9\times \frac{1}{6}+...+17\times \frac{1}{6}=\frac{72}{6}





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