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lbvjy [14]
3 years ago
6

You plan on simultaneously throwing seven fair coins. If you get at least twice as many headsas tails, you’ll stop; otherwise, y

ou’ll throw all seven coins again. How many throws do you expectit will take to stop? [WISE]
Mathematics
1 answer:
Ede4ka [16]3 years ago
7 0

Answer:

the expected value of throws is E(Y) = 15 shots  

Step-by-step explanation:

We will divide the experiment in 2

first) throw seven fair coins , and you get at least twice as many heads as tails when the number of heads is 6 or 7

being the random variable X=getting x heads , then P(X) has a binomial distribution, with p=0.5 and n=7. then

p₂=P(X≥6) =P(X=6)+ P(X=7)= 0.0546 +0.0078 =0.0624

second) if you get at least twice as many heads as tails when the number of heads , you throw again. then the random variable Y= number of shots in the experiment , then Y follows a negative binomial distribution , with probability of success pn=1-p₂ , and number of failures until the experiment is stopped r=1

then the expected value of throws E(Y) is

E(Y) = pn*r/(1-pn) = (1-p₂)*1/p₂ = 1/p₂ -1 = 1/0.0624 - 1 = 15

E(Y) = 15 shots  

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