Question

An urn contains 5 red, 2 white and 3 black balls. Three balls are drawn one by one, at random with replacement. Find the probability distribution of the number of white balls​

Answer 1

$\textbf{Given:}Given:</p><p></p><p>\text{5 red balls, 2 white balls, 3 black balls}5 red balls, 2 white balls, 3 black balls</p><p></p><p>\text{Let X denote the number of white balls when 3 balls are drawn one by one}Let X denote the number of white balls when 3 balls are drawn one by one</p><p></p><p>\text{Then, X can take the values 0, 1 and 2}Then, X can take the values 0, 1 and 2</p><p></p><p>P(X=0)P(X=0)</p><p></p><p>=\displaystyle\frac{8_{C_3}}{10_{C_3}}=10C38C3</p><p></p><p>=\displaystyle\frac{\frac{8{\times}7{\times}6}{1{\times}2{\times}3}}{\frac{10{\times}9{\times}8}{1{\times}2{\times}3}}=1×2×310×9×81×2×38×7×6</p><p></p><p>=\displaystyle\frac{8{\times}7{\times}6}{10{\times}9{\times}8}=10×9×88×7×6</p><p></p><p>=\displaystyle\frac{7}{5{\times}3}=5×37</p><p></p><p>=\displaystyle\frac{7}{15}=157</p><p></p><p>P(X=1)P(X=1)</p><p></p><p>=\displaystyle\frac{8_{C_2}{\times}2_{C_1}}{10_{C_3}}=10C38C2×2C1</p><p></p><p>=\displaystyle\frac{\frac{8{\times}7}{1{\times}2}(2)}{\frac{10{\times}9{\times}8}{1{\times}2{\times}3}}=1×2×310×9×81×28×7(2)</p><p></p><p>=\displaystyle\frac{7}{5{\times}3}=5×37</p><p></p><p>=\displaystyle\frac{7}{15}=157</p><p></p><p>P(X=2)P(X=2)</p><p></p><p>=\displaystyle\frac{8_{C_1}{\times}2_{C_2}}{10_{C_3}}=10C38C1×2C2</p><p></p><p>=\displaystyle\frac{\8{\times}1}{\frac{10{\times}9{\times}8}{1{\times}2{\times}3}}=1×2×310×9×8\8×1</p><p></p><p>=\displaystyle\frac{1}{15}=151</p><p></p><p>\therefore\textbf{The probability distribution of X is}∴The probability distribution of X is</p><p></p><p>\begin{gathered}\begin{array}{|c|c|c|c|}\cline{1-4}x&0&1&2\\\cline{1-4}P(X=x)&\frac{7}{15}&\frac{7}{15}&\frac{1}{15}\\\cline{1-4}\end{array}\end{gathered}\cline1−4x\cline1−4P(X=x)\cline1−4015711572151</p><p></p><p>$

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

➢ Given that,

• An urn contains 5 red, 2 white and 3 black balls.

• Three balls are drawn one by one, at random with replacement.

➢ So, total number of balls = 5 + 2 + 3 = 10

Let p represents the probability of getting a white ball.

and

Let q represents the probability of not getting a white ball.

So,

$\rm :\longmapsto\:p = \dfrac{2}{10} = \dfrac{1}{5}$

and

$\rm :\longmapsto\:q = \dfrac{8}{10} = \dfrac{4}{5}$

We know,

➢ In Binomial Distribution, the probability of random r is given by

$\boxed{ \bf{ \:P(r) = \: ^nC_r \: {p}^{r} {q}^{n - r}}}$

where,

• n is number of independent trials

• p is probability of success

• q is probability of failure

• r is random variable

Thus,

As to find the probability distribution of the number of white balls, random variable r assumes the value 0, 1, 2, 3

So,

P(0) = Probability of getting no white ball

$\rm \:  =  \:  \:^3C_0 \: {\bigg(\dfrac{1}{5} \bigg) }^{0}{\bigg(\dfrac{4}{5} \bigg) }^{3}$

$\rm \:  =  \:  \:1 \times 1 \times \dfrac{64}{125}$

$\rm \:  =  \:  \: \dfrac{64}{125}$

P(1) = Probability of getting 1 white ball

$\rm \:  =  \:  \:^3C_1 \: {\bigg(\dfrac{1}{5} \bigg) }^{1}{\bigg(\dfrac{4}{5} \bigg) }^{3 - 1}$

$\rm \:  =  \:  \:3 \times \dfrac{1}{5} \times \dfrac{16}{25}$

$\rm \:  =  \:  \:\dfrac{48}{125}$

P(2) = Probability of getting 2 white balls

$\rm \:  =  \:  \:^3C_2 \: {\bigg(\dfrac{1}{5} \bigg) }^{2}{\bigg(\dfrac{4}{5} \bigg) }^{3 - 2}$

$\rm \:  =  \:  \:3 \times \dfrac{1}{25} \times \dfrac{4}{5}$

$\rm \:  =  \:  \:\dfrac{12}{125}$

P(3) = Probability of getting 3 white balls = 0

Hence,

➢ Probability distribution table of getting 3 white balls is follow :

$\begin{gathered}\boxed{\begin{array}{c|c} \bf r & \bf P(r) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf \dfrac{64}{125} \\ \\ \sf 1 & \sf \dfrac{48}{125} \\ \\ \sf 2 & \sf \dfrac{12}{125} \\ \\ \sf 3 & \sf 0 \end{array}} \\ \end{gathered}$