An unbiased coin is tossed 4 times. Evaluate: 1) The number of equally likely ou

Question

An unbiased coin is tossed 4 times. Evaluate: 1) The number of equally likely outcomes in the game. 2) P(A), where A is the event that the number of times thecoin comes up head is two. 3) P(B), where B is the event that the number of times thecoin comes up heads at most two times. 4) P(C), where C is the event that the coin comes up head atleast three times. An unbiased coin is tossed 4 times. Evaluate: 1) The number of equally likely outcomes in the game. 2) P(A), where A is the event that the number of times thecoin comes up head is two. 3) P(B), where B is the event that the number of times thecoin comes up heads at most two times. 4) P(C), where C is the event that the coin comes up head atleast three times.

Explanation / Answer

Let X be the number of heads after tossing 4 times. X followsBinomial with n = 4 and p = 0.5. Then, P(X = x) = (nCx)(p^x)(1-p)^(n-x) 1) There are 2 possible outcomes in the first toss, 2 possibleoutcomes in the second toss, etc... Thus, in total, there are 2x2x2x2 = 16 outcomes 2) P(A) = P(X = 2) = (4C2)(0.5)^2(0.5)^2 = 6(0.5)^4 =0.375 Note: 4C2 means "4 chooses 2" 3)P(B) = P(X = 0) + P(X = 1) + P(X = 2) =(4C0) (0.5^0)(0.5^4) + (4C1) (0.5^1)(0.5^3) + 0.375 =0.0625+0.25+0.375 = 0.6875 4)P(C) = P(X=3) + P(X=4) = 1 - P(B) = 1-0.6875 = 0.3125 That is because event C is the complement of event B

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