Suppose that we have 10 coins, which are weighted so thatwhen flipped theith coin shows heads with probabilityp=i/10 (i= 1, . . . ,10). You

Question

Suppose that we have 10 coins, which are weighted so thatwhen flipped theith coin shows heads with probabilityp=i/10 (i= 1, . . . ,10). You randomly pick onecoin, you flip it, and it shows heads.Without putting this coin back, you pick a second coin. What is the probability that the second coin shows heads, given that the first coin showed heads?

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Melody 3 weeks 2022-01-07T04:11:54+00:00 1 Answer 0 views 0

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    2022-01-07T04:13:49+00:00

    Answer:

    the probability that the second coin shows heads, given that the first coin showed heads is 8/15 (5.33%)

    Step-by-step explanation:

    denoting event B= obtaining heads in the fist shot , Ai= probability of obtaining the heads with the i-th coin , and C= probability of obtaining the i-th coin= 1/10 ( assuming that each coin is equally likely to be chosen), then he probability of obtaining heads in the first shot is

    P(B)= ∑ P(Ai)*P(C) for i from 1 to 10

    P(B)= ∑ P(Ai)*P(C) =  ∑ i/10*1/10 = 1/100*∑ i = 1/100* 10*(10+1)/2 = 11/20

    denoting D = obtaining heads in the second shot, then

    P(D∩B)= ∑ probability of choosing the i-th coin*probability of getting heads with the i-th coin * probability of getting heads in the second shot with the remaining coins

    P(D)= ∑ (1/10) * (i/10) * [(∑j/10*1/9)- (1/9)*(i/10)] with i from 1 to 10 and j from 1 to 10

    P(D)= 1/9000 *∑ i * [(∑ j) – i] = 1/9000[(∑ i) *(∑ j) – ∑i²]= 1/9000[(∑ i)² – ∑i²} = 1/9000 *[ [10*(10+1)/2]² – [ 10*(10+1)*(2*10+1)/6] ] = 1/9000 * (25*121 – 10*11*21/6)

    = (25*121 – 5*77)/9000 = 2640/9000 = 22/75

    then from Bayes

    P(D/B)= P(D∩B)/ P(B)= 22/75 /(11/20)= 8/15 (5.33%)

    thus the probability that the second coin shows heads, given that the first coin showed heads is 8/15 (5.33%)

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