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  • Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 28 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 28 maths textbook solution

Answers (1)

Answer:

X 0 1 2 3 4
P\left ( X \right ) \frac{1}{495} \frac{32}{495} \frac{168}{495} \frac{224}{495} \frac{70}{495}


Given: 4 balls are drawn without replacement from box containing 8 red and 4 white balls. If X denotes the no. of red balls drawn, find the probability distribution of X.Hint: Use combination formula

Solution: Let X denote the no of red balls drawn in a random draw of 4 balls

∴X Can take values 0, 1, 2, 3 or 4

Since bag contains 8 red and 6 white balls i.e. total of 12 balls.

∴ Total no. of ways of selecting 4 balls and of 12 =12c_{1} for selecting 0 red balls are will select all 3 balls from red

\therefore P(x=0)=P  (not selecting any red ball)

=\frac{4 c_{4}}{12 c_{4}}=\frac{4 \times 3 \times 2 \times 1}{12 \times 11 \times 10 \times 9}=\frac{1}{495}

For selecting 1 red ball, we will select 1 ball from 8 red and 3 balls from 4 white

\therefore P(x=1)=\frac{4 c_{3} \times 8 c_{1}}{12 c_{4}}=\frac{4 \times 8}{495}=\frac{32}{495}

For selecting 2 red ball, we will select 2 balls from 8 red and 2 balls from 4 white

\therefore P(x=2)=\frac{4 c_{2} \times 8 c_{2}}{12 c_{4}}=6 \times \frac{28}{495}=\frac{168}{495}

For selecting 3 red ball, we will select 3 balls from 8 red and 1 ball from 4 white

\therefore P(x=3)=\frac{4 c_{1} \times 8 c_{3}}{12 c_{4}}=4 \times \frac{56}{495}=\frac{224}{495}

\therefore P(x=4)=P($ selecting all red ball $)

=\frac{8 c_{4}}{12 c_{4}}=\frac{8 \times 7 \times 6 \times 5}{12 \times 11 \times 10 \times 9}=\frac{70}{495}

Now we have P(x) and X

X 0 1 2 3 4
P\left ( X \right ) \frac{1}{495} \frac{32}{495} \frac{168}{495} \frac{224}{495} \frac{70}{495}
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