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Q.There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two-headed coin? OR A die is thrown 6 times. If 'getting an odd number' is a success, then what is the probability of

(i) 5 successes;
(ii) at least 5 successes;
(iii) at most 5 successes?
Meghalaya MboseMBOSE Meghalaya Intermediate Board 2022Subjective· 6mImportance★★★★★
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Set up prior probabilities and likelihoods of a head for each coin, then apply Bayes' theorem.

Let E1,E2,E3E_1, E_2, E_3 be the events that the chosen coin is the two-headed, the biased, and the unbiased coin. Since one is chosen at random,

P(E1)=P(E2)=P(E3)=13.P(E_1) = P(E_2) = P(E_3) = \frac13.

Likelihoods of heads HH:

P(H∣E1)=1,P(H∣E2)=75100=34,P(H∣E3)=12.P(H\mid E_1) = 1, \qquad P(H\mid E_2) = \frac{75}{100} = \frac34, \qquad P(H\mid E_3) = \frac12.

By Bayes' theorem,

P(E1∣H)=P(E1)P(H∣E1)P(E1)P(H∣E1)+P(E2)P(H∣E2)+P(E3)P(H∣E3).P(E_1\mid H) = \frac{P(E_1)P(H\mid E_1)}{P(E_1)P(H\mid E_1) + P(E_2)P(H\mid E_2) + P(E_3)P(H\mid E_3)}.

=13⋅113⋅1+13⋅34+13⋅12=1313(1+34+12)=11+34+12=194=49.= \frac{\tfrac13\cdot 1}{\tfrac13\cdot 1 + \tfrac13\cdot\tfrac34 + \tfrac13\cdot\tfrac12} = \frac{\tfrac13}{\tfrac13\left(1 + \tfrac34 + \tfrac12\right)} = \frac{1}{1 + \tfrac34 + \tfrac12} = \frac{1}{\tfrac{9}{4}} = \frac{4}{9}.

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