Probability of choosing red and black billiard balls

SaqlainSajid

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Hey guys
I have this question about solving a probability math, it’s like this - you have to find the probability of getting two different coloured balls in two choosings from a mixture of 5 red balls and 10 white balls
Now we know
if A and B are two independent events their probability is
P(A ∩ B) = p(A) * p(B) right? If the coin is tossed 2ice
and we get 2 heads, then the probability would be = {H}*{H} = HH dividedby the total no of possible outcomes (sample space)

Don’t the probability of getting a red ball at first and then getting a white ball depend on each other? Cuz if I take a red ball then a total of 14 i.e 4 red coloured and 10 white coloured balls would be left from there we have to choose the next ball, if we had gotten a white ball instead of the red ball the first time, then there’d be 5 red balls and 9 white balls left, so don’t you think the probability of the second choosing depends on the 1st choosing? Then these are not independent events right?

so why have we multiplied 5/15 with 10/14 to find the probability of the 1st ball being red and the second one being white? (5/15*10/14) Since they are not independent, we can’t multiply these to get the probability right?
 

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Although it is not explicitly stated, it is clear that they intend the selections to be "without replacement", so that what you pick first affects what you pick second, and the events are not independent.

What they are doing is using a variation of the multiplication principle, which is best expressed this way:

\(\displaystyle P(A \cap B) = P(A) \cdot P(B | A)\)

That is, to find the probability of two things happening one after the other, you can multiply the probability of the first, by the probability that the second happens, GIVEN that the first has happened. This takes into account the dependency of the second on the first. Note that they multiply 3/15 times 10/14, which is the probability that the second is white, given that the first was red (which changed the numbers). If the events were independent (because the selection was "with replacement"), they would have multiplied by 10/15.

Does that help?
 
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