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A die is thrown three times. Events A and B are defined as below: A : 4 on the third throw B : 6 on the first and 5 on the second throw Find the probability of A given that B has already occurred
Don't you think that the first, second, and the third throws are independent of each other?
If we got six on the first throw, and five on the second, how does that affect the probability of four occurring on the third throw?
1,2, and 3 are different
need to use conditional probablity , i dont know how they affect
They don't. The probability is 1/6.
1/6 for which ?
We're given that B has already occurred. But who cares about what has occurred? We just want to see what the probability of getting a 4 is on the third throw. That is 1/6.
but i need to solve this by conditional probablity
if 4 had occured in 1st or 2nd place then
will it would have been 1/4 as usual
OK, the probability is either 1/6 * 1/6 * 1/6 or just 1/6 depending on the use of "given that B has already occurred".
lol m getting confused
what PK said is correct but if you want to do it conditional probability then here it is P(A/B) = P(A (intersection) B ) / P(B) P(A (intersection) B ) is the event where all the 3 throws have some specified values P(B) is the event where only 2 are specified I hope you can calculate both values
It is 1/6 as A and B are independent. No matter B happened or not, the probability of A happening is the same. Past results of die throws does not affect the future results of die throws. Basically it is what ParthKohli is saying.