From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable denote the number of defective items in the sample. If the variance of is then is equal to

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $σ^{2},$ then $96 σ^{2}$ is equal to _____________ . [2024]

Ans.
(56)

Total number of items = 10

Number of defective items = 3

$∴$ Number of non-defective items = 7

X 0 1 2 3

P(X) $\frac{{}^{7}{C_{5}}}{{}^{10}{C_{5}}} = \frac{1}{12}$ $\frac{{}^{7}{C_{4} \times {}^{3}{C_{1}}}}{{}^{10}{C_{5}}} = \frac{5}{12}$ $\frac{{}^{7}{C_{3} \times {}^{3}{C_{2}}}}{{}^{10}{C_{5}}} = \frac{5}{12}$ $\frac{{}^{7}{C_{2} \times {}^{3}{C_{3}}}}{{}^{10}{C_{5}}} = \frac{1}{12}$

Now, $σ^{2} = \sum {(X - \bar{X})}^{2} P (X)$

$\bar{X} = \sum X P (X) = 0 + \frac{5}{12} + \frac{10}{12} + \frac{3}{12} = \frac{18}{12} = \frac{3}{2}$

$∴$ $σ^{2} = \sum {(X - \frac{3}{2})}^{2} P (X)$

$= \frac{9}{4} \times \frac{1}{12} + \frac{1}{4} \times \frac{5}{12} + \frac{1}{4} \times \frac{5}{12} + \frac{9}{4} \times \frac{1}{12} = \frac{7}{12}$

$\Rightarrow 96 σ^{2} = 8 \times 7 = 56$

Want Us to Email you About Special Offers & Updates?