A balanced dime is tossed twice. The four possible equally likely outcomes are H

Question

A balanced dime is tossed twice. The four possible equally likely outcomes are HH, HT, TH, TT. Let
A=event the first toss is a head
B=event the second toss is a head,
C=event at least one toss is a head
(i) Determine the following probabilities and express your results in words. Compute the conditional probabilities directly without applying the conditional probability rule.
a) P(B) b) P(B|A) c) P(B|C)
d) P(C ) e) P(C|A) f) P(C|(not B))
(ii) Now apply the conditional probability rule to compute the conditional probabilities and compare the results above.

Explanation / Answer

i) can be determined just looking at the sample set - you should obtain the same results ii) P(B) = .5 since there is a 50/50 chance of getting heads and the events are independent P(B|A) = P(A^B)/P(A) = (.5)*(.5)/(.5) = .5 this result is obvious since the events are independent and therefore it doesn't matter what happened before P(B|C) = P(B^C)/P(C) = .5/.75 = (2/3) The probability of B and C I got from looking at the sample set; this is also how I determined P(C) P(C) = .75 P(C|A) = P(C^A)/P(A) = .5/.5 = 1 duh if we know the first coin is heads we know at least one coin is heads P(C|~B) = P(C^~B)/P(~B) = .25/.5 = .5

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