Suppose we have two weighted coins, one of which comes up he

Mary Jackson

Mary Jackson

Answered question

2021-12-16

Suppose we have two weighted coins, one of which comes up heads with probability 0.3, and the other of which comes up heads with probability 0.8. Unfortunately, the coins are otherwise identical, and we have lost track of which is which. Suppose we flip a randomly chosen coin 14 times and let N be the random variable giving the number of heads seen. If in the first 3 flips we see 2 heads, what is the conditional expected number of heads in the 14 flips?
E[N2of first 3 are H]=?

Answer & Explanation

accimaroyalde

accimaroyalde

Beginner2021-12-17Added 29 answers

Step 1
According to given data with the use of binomial distribution:
P(x)=nCxpx(1p)nx
Thus, Probability of coin 1 is:
P(coin 1|2 heads in 3 flips)=P(coin 1 and 2 heads in 3 flips)P(2 heads in 3 flips)
P(coin 1|2 heads in 3 flips)=3C2(0.3)2(0.7)3C2(0.8)2(0.2)+3C2(0.3)2(0.7)
P(coin 1|2 heads in 3 flips)=0.1890.573
P(coin 1|2 heads in 3 flips)=0.3298
Step 2
The probability of coin 2 is:
P(coin 2|2 heads in 3 flips)=P(coin 2 and 2 heads in 3 flips)P(2 heads in 3 flips)
P(coin 2|2 heads in 3 flips)=3C2(0.8)2(0.2)3C2(0.8)2(0.2)+3C2(0.3)2(0.7)
P(coin 2|2 heads in 3 flips)=0.3840.384+0.189
P(coin 2|2 heads in 3 flips)=0.3840.573
P(coin 2|2 heads in 3 flips)=0.6701
Now the probability of head is:
P(Heads)=(0.3×0.3298)+(0.8×0.6701)
P(Heads)=0.0989+0.5360
P(Heads)=0.6349
So E(Y)=11×0.6349=6.98
E(X)=2+6.98=8.98
Lynne Trussell

Lynne Trussell

Beginner2021-12-18Added 32 answers

Step 1
Let H1 be the event that we get a head on the first flip of the chosen coin, and let H2 denote the event that we get a head on the second flip of the chosen coin.
P(H1)=P(H1fair)+P(H1biased)
=P(H1fair)P(fair)+P(H1biased)P(biased)
=12×12+45×12=14+25=1320
Now, the probability of getting two heads in two consecutive flips of the chosen coin is
P(HH)=P(HHFair)+P(HHBiased)=P(HHFair)P(Fair)+P(HHBiased)P(Biased)
=14×12+(45)2×12=18+825=89200
So,
P(H2H1)=P(H1H2)P(H1)=P(HH)P(H1)=892001320=89130
nick1337

nick1337

Expert2021-12-28Added 777 answers

Everything is easy if you count it logically: the expected number of eagles seen with 12 tosses is 8 eagles seen

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