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So I have this problem:

  • First box has 5 white and 4 black balls
  • Second box has 6 white and 3 black balls
  • Third box has one random ball from each box

What is the chance to have 2 white balls in the third box? ( my answer is approximately 37% but I am quite unsure )

2 Answers 2

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Case 1-

From first box white ball is picked. We have,

$\frac{5}{5+4} = \frac 59$

Case 2-

From second box white ball is picked. We have,

$\frac{6}{6+3} = \frac 69$

Probability of both whites in third box

$ = \frac 59 × \frac 69 = \frac {30}{81}$ = 0.37 or 37%

  • 0
    I have additional question. Let's say we take out one ball and it is white. Then we calculate again the chance for the second to be white? Is it different? – 2017-02-01
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    If without replacement then its different. If with replacement then same. – 2017-02-01
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    What is it like without replacement? – 2017-02-01
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    Suppose you have 4 white and 5 red. Then probability of one ball drawn without replacement to be white is $\frac49$. Without replacement so you left with 3 white and 5 red. One another ball is picked. Then probability of white is $\frac38$. Not same. – 2017-02-01
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    Suppose you have 4 white and 5 red. Then probability of one ball drawn with replacement to be white is $\frac49$. With replacement so you have again 4 white and 5 red. One another ball is picked. Then probability of white is $\frac49$. Same. – 2017-02-01
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    In the current problem drawing first white ball would be 11/18 and second 10/17 (no replacement). Correct? – 2017-02-01
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    Yes if you have 11 white and 7 black balls. And I have written red instead of black in my above comments. – 2017-02-01
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$\frac{5}{5+4}$ is the probability to draw a white ball from the first box

$\frac{6}{6+3}$ is the probability to draw a white ball from the seond box

So the probability to have two white in the third box is $\frac{5}{5+4}\frac{6}{6+3} = \frac{30}{81} \approx 0.37$