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mathmath333
 one year ago
Probablity Question
mathmath333
 one year ago
Probablity Question

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0A 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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Don't you think that the first, second, and the third throws are independent of each other?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2If 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?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.01,2, and 3 are different

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0need to use conditional probablity , i dont know how they affect

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2They don't. The probability is 1/6.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2We'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.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0but i need to solve this by conditional probablity

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0if 4 had occured in 1st or 2nd place then

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0will it would have been 1/4 as usual

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2OK, 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".

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0lol m getting confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what 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

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1It 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.
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