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Question
A coin is biased so that the head is 3 times as likely to occur as tail. Find the probability distribution of number of tails in two tosses.
Solution
Let X denote the number of tails.
∴ Possible values of X are 0, 1, 2.
Let P(getting tail) = p
According to the given condition,
P(getting head) = q = 3p
As p + q = 1,
p + 3p = 1
∴ p = `(1)/(4) "and " "q" = (3)/(4)`
∴ P(X = 0) = P(no tails) = qq = q2 = `(3/4)^2 = (9)/(16)`
P(X = 1) = P(one tail) = pq + qp = 2pq = `2(1/4)(3/4) = (6)/(16)`
P(X = 2) = P(two tails) = pp = p2 = `(1/4)^2 = (1)/(6)`
∴ Probability distribution of X is as follows:
| X | 0 | 1 | 2 |
| P(X = x) | `(9)/(16)` | `(6)/(16)` | `(1)/(16)` |
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