74 views
Consider an urn containing $12$ balls, of which $8$ are white.A sample of size $4$ is to be drawn without replacement.What is the conditional probability that the first and third balls drawn will be white given that the sample drawn contains exactly $3$ white balls?
edited | 74 views
0
I think it'll be $\dfrac{1}{2}$
0
yah I am also getting the same.
my approach is ((8/12 * 4/11 * 7/10 * 6/9)+(8/12 * 7/11 * 6/10 * 4/9))  /    ((8c3 * 4)/12c4)
0
With replacement = $\dfrac{1}{2}$

without replacement = $\dfrac{1}{2}$

+1 vote
$\color{red}{\text{Without Replacement}}$

$\color{maroon}{\text{Conditional Probability} = \dfrac{\text{Number of Favourable Ways}}{\text{Total Number of Ways}}}$

Now, the urn having a total of $12$ balls in which $8$ balls are $\color{grey}{white}$.

So, remaining $(12-8)=4$ balls are $\color{black}{black}$.

Given that,

The sample space contains exactly $3$ $\color{grey}{\text{white balls}}$

& the favourable condition is - when the $1^{st}$ and $3^{rd}$ balls are of $\color{grey}{\text{white colour}}$.

Sample space contains exactly $3$ $\color{grey}{\text{white balls}}$

$S_1 = \{(\color{grey}{WWW}\color{black}{B},\color{grey}{WW}B\color{grey}{W},\color{grey}{W}B\color{grey}{WW},B\color{grey}{WWW})\}$

Sample space having $1^{st}$ and $3^{rd}$ balls as $\color{grey}{white}$.

$S_2=\{(\color{grey}{W}B\color{grey}{W}B),(\color{grey}{W}B\color{grey}{WW}),(\color{grey}{WWWW}),(\color{grey}{WWW}B)\}$

$∴S_{1}S_{2} = \{(\color{grey}{W}B\color{grey}{WW}),(\color{grey}{WWW}B)\}$

$∴ P(S_1) = \dfrac{8}{12}\times\dfrac{7}{11}\times\dfrac{6}{10}\times\dfrac{4}{9}+\dfrac{8}{12}\times\dfrac{7}{11}\times\dfrac{4}{10}\times\dfrac{6}{9}+\dfrac{8}{12}\times\dfrac{4}{11}\times\dfrac{7}{10}\times\dfrac{6}{9}+\dfrac{4}{12}\times\dfrac{8}{11}\times\dfrac{7}{10}\times\dfrac{6}{9}$

= $4\times\dfrac{4\times8\times7\times6}{12\times11\times10\times9}$

$∴ P(S_2) = \dfrac{8}{12}\times\dfrac{4}{11}\times\dfrac{7}{10}\times\dfrac{3}{9}+\dfrac{8}{12}\times\dfrac{4}{11}\times\dfrac{7}{10}\times\dfrac{6}{9}+\dfrac{8}{12}\times\dfrac{7}{11}\times\dfrac{6}{10}\times\dfrac{5}{9}+\dfrac{8}{12}\times\dfrac{7}{11}\times\dfrac{6}{10}\times\dfrac{4}{9}$

$∴ P(S_{1}S_{2}) = \dfrac{8}{12}\times\dfrac{4}{11}\times\dfrac{7}{10}\times\dfrac{6}{9} + \dfrac{8}{12}\times\dfrac{7}{11}\times\dfrac{6}{10}\times\dfrac{4}{9}$

= $2\times\dfrac{8\times4\times7\times6}{12\times11\times10\times9}$

$∴ \textbf{Required Probability} =$ $\dfrac{P(S_{1}S_{2})}{P(S_{1})} = \dfrac{2\times\dfrac{8\times4\times7\times6}{12\times11\times10\times9}}{4\times\dfrac{4\times8\times7\times6}{12\times11\times10\times9}}$

= $\dfrac{2}{4}$

=  $\dfrac{1}{2}$
edited
+1 vote
$\color{red}{\text{With Replacement}}$

$\color{maroon}{\text{Conditional Probability} = \dfrac{\text{Number of Favourable Ways}}{\text{Total Number of Ways}}}$

Now, the urn having a total of $12$ balls in which $8$ balls are $\color{grey}{white}$.

So, remaining $(12-8)=4$ balls are $\color{black}{black}$.

Given that,

The sample space contains exactly $3$ $\color{grey}{\text{white balls}}$

& the favourable condition is - when the $1^{st}$ and $3^{rd}$ balls are of $\color{grey}{\text{white colour}}$.

Sample space contains exactly $3$ $\color{grey}{\text{white balls}}$

$S_1 = \{(\color{grey}{WWW}\color{black}{B},\color{grey}{WW}B\color{grey}{W},\color{grey}{W}B\color{grey}{WW},B\color{grey}{WWW})\}$

Sample space having $1^{st}$ and $3^{rd}$ balls as $\color{grey}{white}$.

$S_2=\{(\color{grey}{W}B\color{grey}{W}B),(\color{grey}{W}B\color{grey}{WW}),(\color{grey}{WWWW}),(\color{grey}{WWW}B)\}$

$∴S_{1}S_{2} = \{(\color{grey}{W}B\color{grey}{WW}),(\color{grey}{WWW}B)\}$

$∴ P(S_1) = 4 \times (\dfrac{4}{12}\times\dfrac{8}{12}\times\dfrac{8}{12}\times\dfrac{8}{12})$

$∴ P(S_2) = (\dfrac{8}{12}\times\dfrac{8}{12}\times\dfrac{4}{12}\times\dfrac{4}{12}) + 2\times(\dfrac{8}{12}\times\dfrac{8}{12}\times\dfrac{8}{12}\times\dfrac{4}{12}) + (\dfrac{8}{12}\times\dfrac{8}{12}\times\dfrac{8}{12}\times\dfrac{8}{12})$

$∴ P(S_{1}S_{2}) = 2 \times (\dfrac{8}{12}\times\dfrac{8}{12}\times\dfrac{8}{12}\times\dfrac{4}{12})$

∴ Required Probability = $\dfrac{P(S_{1}S_{2})}{P(S_{1})} = \dfrac{2\times\dfrac{8}{12}\times\dfrac{8}{12}\times\dfrac{8}{12}\times\dfrac{4}{12}}{4\times\dfrac{4}{12}\times\dfrac{8}{12}\times\dfrac{8}{12}\times\dfrac{8}{12}}$

= $\dfrac{2\times8\times8\times8\times4}{4\times4\times8\times8\times8}$

=  $\dfrac{1}{2}$
edited
0
sorry i actually wrongly typed the question.It will be the experiment for without replacement.I edited the question.For with replacement your solution is perfect.Please solve it for without replacement.
0
Ok.. I'm giving

that'll also be $\dfrac{1}{2}$
0
yes.I also getting same result.I mentioned my approach in comment.

1
2
+1 vote
3
4
6
+1 vote