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briozeal
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briozeal
Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, What is the probability that Kate has more than $10 but less than $15 ?

OA will follow !


problem = probability of having 1 to 4 tails

5C1(1/32) + 5C2(1/32)...5C4(1/32)

(1/32)*(5 + 5*4/2 + 5*4*3/6 + 5) = 30/32 = 15/16
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ggarr
Quote:
Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, What is the probability that Kate has more than $10 but less than $15 ?
Given the above, Kate can only win either 11 or 13 dollars. Therefore, she needs to win either 3 or 4 flips:

5C3 = 10
5C4 = 5

There are a total of 32 different outcomes:
TTTTT
FFFFF
5C1
5C2
5C3
5C4

Probability of the above happening is 15/32



hah ! silly of me ... 5C1 & 5C2 dont qualify ! thank you
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where does it say that Kate has to win? 15>kate>10?

1 heads--5C1
2 heads--5C2
3 heads--5C3
4 heads--5C4
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adub35
where does it say that Kate has to win? 15>kate>10?

1 heads--5C1
2 heads--5C2
3 heads--5C3
4 heads--5C4


If one person wins, the other loses. If Katie wins once, Danny wins 4 times and Katie wins $1, but loses $4.

I did it the same way. I said that each coin flip has 2 outcomes, and there are 5 flips so its 2^5=32. I wrote out HHHTT and HHHHT as the only ways Katie can win. 3 heads is (5 c 3) and 4 heads is (5 c 4). So its (10 + 5)/32 =15/32



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