Explain answers for 10 points
a) A gambler has two coins, A and B; tbe probabilities of their turning up heads are 0.8 and 0.4 respectively. One coin is selected at random and is tossed twice, and a head and a tail are observed. Find the probability that the coin selected was A.
a) A gambler has two coins, A and B; tbe probabilities of their turning up heads are 0.8 and 0.4 respectively. One coin is selected at random and is tossed twice, and a head and a tail are observed. Find the probability that the coin selected was A.
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If he chose coin A, then:
p(one head, one tail)
= p(head then tail) + p(tail then head)
= 0.8 * 0.2 + 0.2 * 0.8
= 0.32
If he chose coin B, then:
p(one head, one tail)
= p(head then tail) + p(tail then head)
= 0.6 * 0.4 + 0.4 * 0.6
= 0.48
From here there are two ways to solve it. The intuitive way, and the way with formulas.
The intuitive way:
Out of every 200 throws, we get each coin 100 times. So we get 1 head and 1 tail 32 times due to coin A, and 48 times due to coin B. So if we see 1 head and 1 tail, the probability we chose coin A is 32 / (32 + 48) = 32/80 = 2/5 = 0.4
The formula way:
Let C be event that we get 1 head 1 tail. Its overall probability
p(C) = p(B)p(C|B) + p(A)p(C|A) = 0.5 * 0.48 + 0.5 * 0.32 = 0.4
p(C and A) = p(A) p(C | A) = 0.5 * 0.32 = 0.16
but also:
p(C and A) = p(C) p(A | C)
=> p(A | C) = p(C and A) / p(C) = 0.16 / 0.4 = 0.4
p(one head, one tail)
= p(head then tail) + p(tail then head)
= 0.8 * 0.2 + 0.2 * 0.8
= 0.32
If he chose coin B, then:
p(one head, one tail)
= p(head then tail) + p(tail then head)
= 0.6 * 0.4 + 0.4 * 0.6
= 0.48
From here there are two ways to solve it. The intuitive way, and the way with formulas.
The intuitive way:
Out of every 200 throws, we get each coin 100 times. So we get 1 head and 1 tail 32 times due to coin A, and 48 times due to coin B. So if we see 1 head and 1 tail, the probability we chose coin A is 32 / (32 + 48) = 32/80 = 2/5 = 0.4
The formula way:
Let C be event that we get 1 head 1 tail. Its overall probability
p(C) = p(B)p(C|B) + p(A)p(C|A) = 0.5 * 0.48 + 0.5 * 0.32 = 0.4
p(C and A) = p(A) p(C | A) = 0.5 * 0.32 = 0.16
but also:
p(C and A) = p(C) p(A | C)
=> p(A | C) = p(C and A) / p(C) = 0.16 / 0.4 = 0.4
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one each of a head & a tail can appear either as HT or TH
also, each coin has a 0.5 probability of being selected
P[coin A & one each head-tail] = 0.5*2*0.8*0.2 = 0.16
P[coin B & one each head-tail] = 0.5*2*0.4*0.6 = 0.24
P[one each of head-tail] = sum of the above = 0.40
P[coin A given one each head-tail] = 0.16 / 0.40 = 0.4 <-----------
also, each coin has a 0.5 probability of being selected
P[coin A & one each head-tail] = 0.5*2*0.8*0.2 = 0.16
P[coin B & one each head-tail] = 0.5*2*0.4*0.6 = 0.24
P[one each of head-tail] = sum of the above = 0.40
P[coin A given one each head-tail] = 0.16 / 0.40 = 0.4 <-----------