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Probability Distributions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability Distributions3 Marks
Question
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers. Find E(X).
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Answer

Two numbers are chosen from the first 6 positive integers.
$\therefore n ( S )={ }^6 C_2=\frac{6 \times 5}{1 \times 2}=15$
Let $X$ denote the larger of the two numbers.
Then $X$ can take values $2,3,4,5,6$.
When $X=2$, the other positive number which is less than $2$ is $1.$
$ \therefore n ( X )=1$
$\therefore P ( X =2)= P (2)=\frac{n(X)}{n(S)}=\frac{1}{15} $
When $X=3$, the other positive number less than 3 can be 1 or 2 and hence can be chosen in 2 ways.
$ \therefore n ( X )=2$
$P ( X =3)= P (3)=\frac{n(X)}{n(S)}=\frac{2}{15} $
Similarly, $P(X=4)=P(4)=\frac{3}{15}$
$ P(X=5)=P(5)=\frac{4}{15}$
$P(X=6)=P(6)=\frac{5}{15}$
$\therefore E(X)=\sum x_i P\left(x_i\right)$
$=2 \times \frac{1}{15}+3 \times \frac{2}{15}+4 \times \frac{3}{15}+5 \times \frac{4}{15}+6 \times \frac{5}{15}$
$=\frac{2+6+12+20+30}{15}$
$=\frac{70}{15}$
$=\frac{14}{3} $

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