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?

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%)

i can answer this question sweetie!

step-by-step explanation:

your answer would be b

step-by-step explanation: