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Problem: We roll a die. If we obtain a 6, we choose a ball from box A where three balls are white and two are black. If we do not obtain a 6, we choose a ball from box B where two balls are white and four are black.

(a) What is the probability of obtaining a white ball?

The answer on my notes went something like this: $P[\text{white ball}] = P[\text{get 6} \cap \text{white ball}]$ + $P[\text{not 6} \cap \text{whiteball}]$.

Why not $P[\text{white ball}] = P[\text{white ball} | \text{get 6}]$ + $P[\text{white ball} | \text{not 6}]$?

What's the difference here?

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  • $\begingroup$ Welcome to CV. Please refrain from posting text as pictures, as this can be inconvenient for mobile users and screenreaders. $\endgroup$ – Frans Rodenburg Oct 6 at 1:14
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    $\begingroup$ Please include the self study tag. $\endgroup$ – Michael R. Chernick Oct 6 at 1:56
  • $\begingroup$ $P(W) = P(W \cap 6) + P(W \cap 6^c) = P(6)P(W|6) + P(6^c)P(W|6^c) = \cdots = ?$ Can you give a justification for each step and then finish with a numerical answer? $\endgroup$ – BruceET Oct 6 at 2:26
  • $\begingroup$ @MrPi (and to who approved the edit): The self-study tag is to be added by the OP, as to encourage new members with homework related questions to read its description. Otherwise Michael wouldn't have bothered commenting about it and simply would have added it himself. $\endgroup$ – Frans Rodenburg Oct 10 at 6:44
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$P(\text{white ball}|\text{get 6}) + P(\text{white ball}|\text{not 6})$ just summates the probability of obtaining a white ball in the scenario you obtained six with the probability of obtaining a white ball in the scenario you did not obtain six.

You can see that this expression is incorrect by two ways. First, it is not a valid probability. For example, if the two boxes had only white balls, this expression would be 2. You might suggest to solve this issue by replacing the summation with averaging: $\frac{1}{2}P(\text{white ball}|\text{get 6}] + \frac{1}{2}P(\text{white ball}|\text{not 6})$

However, here comes the second issue with this expression - you must take into account $P(\text{get 6})$ and $P(\text{not 6})$. if you just average the conditional probabilities, you implicitly assume that $P(\text{get 6})$=$P(\text{not 6})=0.5$.

The correct solution using conditional probabilities requires multiplying them with the prior probabilities: $P(\text{white ball}|\text{get 6})P(\text{get 6}) + P(\text{white ball}|\text{not 6}]P(\text{not 6})$. This solution is equal to your first solution since $P(A \cap B)=P(A|B)P(B)$.

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