Two coins are tossed independently where P(head occurs when coin i is tossed) = $p_i$ , $i = 1, 2$. Given that at least one head has occurred, the probability that coins produced different outcomes is
(A)$\frac{2p_1p_2}{p_1+p_2-2p_1p_2}$ (B) $\frac{p_1+p_2-2p_1p_2}{p_1+p_2-p_1p_2}$ (C) $\frac{2}{3}$ (D) none of the above
Possible Outcomes: 3(HT,TH,HH). Required Outcomes: 2(HT,TH). So, is the answer $\frac{2}{3}$?
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$\begingroup$Answer:
$$P(H_1) = p_1, P(H_2) = p_2; P(T_1) = 1-p_1 P(T_2) = 1-p_2$$
Since the coin tosses are independent, $$P(H_1 T_2) = p_1(1-p_2)$$ $$P(H_1 H_2) = p_1p_2$$ $$P(T_1 T_2) = (1-p_1)(1-p_2)$$ $$P(T_1 H_2) =(1- p_1)p_2$$
P(different outcomes) $$= p_1(1-p_2)+(1- p_1)p_2 .... (1)$$ P(Head appearing in one of the coins) $$= p_1(1-p_2)+(1- p_1)p_2+p_1p_2 .... (2)$$
Now the conditional probability P(different outcomes/One head has appeared) =
$$ \dfrac{p_1+p_2-2p_1p_2}{p_1+p_2-p_1p_2}$$
That will be your answer
Thanks
Satish
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