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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 18 maths textbook solution[

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 18 maths textbook solution

Answers (1)

Answer:

X 0 1 2 3
P\left ( X \right ) \frac{1}{6} \frac{1}{2} \frac{3}{10} \frac{1}{30}


Hint: use combination formula

Given: find the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement from a bag containing 4 white and 6 red balls.

Solution: Let x denote the number balls drawn in a random draw of 3 balls

∴x can take values 0, 1, 2 or 3

Since bag contains 6 red and 4 white ball, i.e.  total of 10 balls.

∴ Total no of ways of selecting 3 balls put of 10 = 10c_{3}

For selecting 0 white balls, we will select all 3 balls from red

\therefore P\left ( x=0 \right )=P(not selecting any white ball)

=\frac{6 c_{3}}{10 c_{3}}=\frac{6 \times 5 \times 4}{10 \times 9 \times 8}=\frac{1}{6}

For selecting 1 white ball we will select all 2 balls from red and 1 from white

\therefore P(x=1)=\frac{6 c_{2} \times 4 c_{1}}{10 c_{3}}=\frac{15}{30}=\frac{1}{2}

For selecting 2 white balls, we will select and 1 ball from red and 2 from white

\therefore P(x=2)=\frac{6 c_{1} \times 4 c_{2}}{10 c_{3}}=\frac{9}{30}=\frac{3}{10}

For selecting 3 white balls, we will select all 0 balls from red and 3 from white

\therefore P(x=3)=\frac{6 c_{0} \times 4 c_{3}}{10 c_{3}}=\frac{1}{30}

Now we have X and P(X)

∴ Now we have are ready to write the probability distribution of x ÷

The following table gives probability Distributions:

X 0 1 2 3
P\left ( X \right ) \frac{1}{6} \frac{1}{2} \frac{3}{10} \frac{1}{30}
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