learningizk00l
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1.Fair four sided dice with faces labelled 1,2,3 and 4 are rolled one at a time until a 4 is scored
given that a four is not rolled in the first six attempts , find the probability that a four is rolled within the next two attempts.
I tried (3/4)^6×((3/4)+(3/4x1/4))
but the answer is 0.4375

2. Carrie is playing a game of Solitaire. The probability that she wins each game is 0.2.Carrie plays the game until she wins
find the variance of the number of games Carrie plays. This is (1-p)/p^2 which is 0.8/0.2^2 =20 but i don't really understand why could someone please explain why this is how you calculate the variance in this case.
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ry7xsfa
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(Original post by learningizk00l)
1.Fair four sided dice with faces labelled 1,2,3 and 4 are rolled one at a time until a 4 is scored
given that a four is not rolled in the first six attempts , find the probability that a four is rolled within the next two attempts.
I tried (3/4)^6×((3/4)+(3/4x1/4))
but the answer is 0.4375

2. Carrie is playing a game of Solitaire. The probability that she wins each game is 0.2.Carrie plays the game until she wins
find the variance of the number of games Carrie plays. This is (1-p)/p^2 which is 0.8/0.2^2 =20 but i don't really understand why could someone please explain why this is how you calculate the variance in this case.
Are you sure about the answer to question 1? Maybe it's worth asking a teacher.

$0.4375 = 0.25 + (0.75 * 0.25)$

This makes sense as P(X = 1) OR P(X = 2), and may be used because the rolls are independent, but it really seems to me that you should be looking for P(7\leq X \leq 8), which is what you have done.

In terms of your second question, I can't tell you the reasons why, but think about your modelling. She's playing a game where the probability of success is constant, and each game is independent. She's also playing until she has a single success. This means it makes sense to model this as a Geometric Distribution with probability 0.2.

It therefore comes that the variance of this is \frac{(1-p)}{p^2} as this is how we calculate the variance of such a distribution
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