probability question
can someone help me with approaching this question?all I have done is 5!/3!2! to find the combinations, but idk if this is right
(edited 12 months ago)
Original post by lavely
can someone help me with approaching this question?all I have done is 5!/3!2! to find the combinations, but idk if this is right
Sounds ok. So how many combination would have a last disc being black (so the combinations of the previous 4). Note as the last disc is either black or white, you may suspect what the answer should be, the "write down" in the question.
(edited 12 months ago)
considering the previous 4 can you do 2 x 1 x 2 x 1 to find the combination of 2 blacks and 2 whites?
Original post by mqb2766
Sounds ok. So how many combination would have a last disc being black (so the combinations of the previous 4). Note as the last disc is either black or white, you may suspect what the answer should be, the "write down" in the question.
Original post by lavely
considering the previous 4 can you do 2 x 1 x 2 x 1 to find the combination of 2 blacks and 2 whites?
It would be similar to your 5!/(3!2!) so number of combinations but account for the double counting.
Original post by lavely
can someone help me with approaching this question?all I have done is 5!/3!2! to find the combinations, but idk if this is right
On a purely pragmatic note: the number of combinations is pretty small. You can write them all down, count how many satisfy the conditions for each part and be done in less than 5 minutes.
Original post by lavely
considering the previous 4 can you do 2 x 1 x 2 x 1 to find the combination of 2 blacks and 2 whites?
Guessing you have the answer? A bit of thought where those 4 (2 white, 2 black) go in the 5 should convince you that the probability of choosing black at any position is the same, so it must be the same as choosing it for the first position. It is after all a 1 marker.
(edited 12 months ago)
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