Question 940217: Please help ! One enterprising gambler decides to take a bet that if he flips a coin 7 times in a row and rolls a die 7 times in a row, he will get exactly 3 heads or 2 sixes. What are his odds of winning the bet ? (answer to the nearest hundredth percent )
Found 3 solutions by ewatrrr, stanbon, mathmate:
Answer by ewatrrr(24785)   (Show Source):
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! One enterprising gambler decides to take a bet that if he flips a coin 7 times in a row and rolls a die 7 times in a row, he will get exactly 3 heads or 2 sixes. What are his odds of winning the bet ? (answer to the nearest hundredth percent )
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P(3 heads in 7 rolls) = 7C3(1/2)^7 = 0.2734
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P(2 sixes) = 7C2(1/6)^2(5/6)^5 = 0.2344
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P(win) = 0.2734+0.2344 = 0.5078
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P(not win) = 1-0.5078 = 0.4922
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Ans: Odds of winning = P(win)/P(not winning) :::: 0.5078/0.4922
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Cheers,
Stan H.
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Answer by mathmate(429)  (Show Source):
You can put this solution on YOUR website! Given:
flips (fair) coin 7 times.
rolls (fair) die 7 times.
Find
Probability of getting 3 heads or 2 sixes or both.
Using binomial distribution,
For coin, p=0.5, n=7, r=3, C(n,r)=n!/((n-r)!r!)
P(3H) = 
For die, p=1/6, n=7, r=2,
P(2-"6") 
Since the two processes can be considered independent, so P(A & B)=P(A)*P(B),
Using
P(A or B) = P(A) + P(B) -P(A & B)
= 0.27344+0.23443-(0.27344*0.23443)
= 0.4438
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