I've gotten up to c, and am confused as to why P(s=5) is 81/256 and not 81/1024??

Question:

Stuart rolls a biased dice. P(six) = 0.25. He stops when he rolls a six and he rolls the dice a maximum of five times. The random variable S represents the number of times he rolls the dice.

c. Construct a table giving the probability distribution of S.

[The values are when s=1,2,3,4,5. 1/4, 3/16,9/64,27/256,81/256.

Question:

Stuart rolls a biased dice. P(six) = 0.25. He stops when he rolls a six and he rolls the dice a maximum of five times. The random variable S represents the number of times he rolls the dice.

c. Construct a table giving the probability distribution of S.

[The values are when s=1,2,3,4,5. 1/4, 3/16,9/64,27/256,81/256.

P(s=5) has to equal 81/256 because remember the sum of all the probabilities has to be 1.

Original post by n-marufu

I've gotten up to c, and am confused as to why P(s=5) is 81/256 and not 81/1024??

Question:

Stuart rolls a biased dice. P(six) = 0.25. He stops when he rolls a six and he rolls the dice a maximum of five times. The random variable S represents the number of times he rolls the dice.

c. Construct a table giving the probability distribution of S.

[The values are when s=1,2,3,4,5. 1/4, 3/16,9/64,27/256,81/256.

Question:

Stuart rolls a biased dice. P(six) = 0.25. He stops when he rolls a six and he rolls the dice a maximum of five times. The random variable S represents the number of times he rolls the dice.

c. Construct a table giving the probability distribution of S.

[The values are when s=1,2,3,4,5. 1/4, 3/16,9/64,27/256,81/256.

the previous poster is correct, if you want to work it out in a bit more detail find P( 5 failures in a row ) + P( 4 failures followed by 1 success ).

Success means getting a 6, failure means not getting a 6

(edited 2 months ago)

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