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This question has been bugging me for a long time and thus resorted to ask it here. Let us assume that there are $5$ balls in a box out of which $3$ are white while the rest $2$ are black. Also, it is given that the balls are distinguishable and we are asked to compute the probability that if we draw two balls, then both the balls will be white.

In this manner the question is pretty easy a basic combination formula will yield the answer. Now, let us assume that the balls are indistinguishable. The total number of cases can still be easily computed with the help of multinomial theorem but how do I calculate the favorable cases?

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  • $\begingroup$ You are only asked to look at the color of the balls after drawing them ("are they both white?") and not at the balls themselves. So distinguishable in any other way is not relevant. $\endgroup$
    – drhab
    Jul 16, 2019 at 15:13
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    $\begingroup$ An alternative to the drhab answer (+1) is : (i) Assume order does not matter we get $$ \frac{{3 \choose 2}}{{5 \choose 2}} = 3/10$$ (ii) Assume order matters $$ \frac{(3)(2)}{(5)(4)} = 3/10$$ $\endgroup$
    – Michael
    Jul 16, 2019 at 15:15

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I preassume in this answer that drawn balls are not replaced.

In both cases (distinguishable balls or not) the answer can be formulated as:$$P(F\cap S)=P(F)P(S\mid F)=\frac35\frac24=\frac3{10}$$where $F$ denotes the event that the first ball drawn is white and $S$ denotes the event that the second ball drawn is white.

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