Question

an urn contains 5 red ,2 white,3 black balls.three balls are drawn one by one at random without replacement . find the probability distribution of the number of white balls.

Answer 1

$\textbf{Given:}$

$\text{5 red balls, 2 white balls, 3 black balls}$

$\text{Let X denote the number of white balls when 3 balls are drawn one by one}$

$\text{Then, X can take the values 0, 1 and 2}$

$P(X=0)$

$=\displaystyle\frac{8_{C_3}}{10_{C_3}}$

$=\displaystyle\frac{\frac{8{\times}7{\times}6}{1{\times}2{\times}3}}{\frac{10{\times}9{\times}8}{1{\times}2{\times}3}}$

$=\displaystyle\frac{8{\times}7{\times}6}{10{\times}9{\times}8}$

$=\displaystyle\frac{7}{5{\times}3}$

$=\displaystyle\frac{7}{15}$

$P(X=1)$

$=\displaystyle\frac{8_{C_2}{\times}2_{C_1}}{10_{C_3}}$

$=\displaystyle\frac{\frac{8{\times}7}{1{\times}2}(2)}{\frac{10{\times}9{\times}8}{1{\times}2{\times}3}}$

$=\displaystyle\frac{7}{5{\times}3}$

$=\displaystyle\frac{7}{15}$

$P(X=2)$

$=\displaystyle\frac{8_{C_1}{\times}2_{C_2}}{10_{C_3}}$

$=\displaystyle\frac{\8{\times}1}{\frac{10{\times}9{\times}8}{1{\times}2{\times}3}}$

$=\displaystyle\frac{1}{15}$

$\therefore\textbf{The probability distribution of X is}$

$\begin{array}{|c|c|c|c|}\cline{1-4}x&0&1&2\\\cline{1-4}P(X=x)&\frac{7}{15}&\frac{7}{15}&\frac{1}{15}\\\cline{1-4}\end{array}$

Answer 2

Answer:

$\textbf{Given:}$

:

$\text{5 red balls, 2 white balls, 3 black balls}5$

$\blue{\text{Let X denote the number of white balls when 3 balls are drawn one by one}}$

$\text{Then, X can take the values 0, 1 and 2}$

$\red{P(X=0)}$

$=\displaystyle\frac{8_{C_3}}{10_{C_3}}$

$=\displaystyle\frac{\frac{8{\times}7{\times}6}{1{\times}2{\times}3}}{\frac{10{\times}9{\times}8}{1{\times}2{\times}3}}$

$=\displaystyle\frac{8{\times}7{\times}6}{10{\times}9{\times}8}$

$=\displaystyle\frac{7}{5{\times}3}$

$=\displaystyle\frac{7}{15}$

$\pink{P(X=1)}$

$</p><p></p><p>=\displaystyle\frac{8_{C_2}{\times}2_{C_1}}{10_{C_3}}$

$=\displaystyle\frac{\frac{8{\times}7}{1{\times}2}(2)}{\frac{10{\times}9{\times}8}{1{\times}2{\times}3}}$

$=\displaystyle\frac{7}{5{\times}3}$

$=\displaystyle\frac{7}{15}$

$\green{P(X=2)}$

$=\displaystyle\frac{8_{C_1}{\times}2_{C_2}}{10_{C_3}}$

$=\displaystyle\frac{8{\times}1}{\frac{10{\times}9{\times}8}{1{\times}2{\times}3}}$

$=\displaystyle\frac{1}{15}$

$\color{cyan}\therefore\textbf{The probability distribution of X is}$

$\begin{gathered}\begin{array}{|c|c|c|c|}\cline{1-4}x&0&1&2\\\cline{1-4}P(X=x)&\frac{7}{15}&\frac{7}{15}&\frac{1}{15}\\{\cline}{1-4}\end{array}\end{gathered} </p><p>1−4x</p><p>1−4P(X=x)</p><p>1−4$

Step-by-step explanation:

pray for me to come over