Maths · Probability: Probability of an event, addition and multiplication theorems of probability
Assertion A fair coin is being tossed four times. Consider the following events:
Assertion A fair coin is being tossed four times. Consider the following events: A is the event all four results are the same. B is the event exactly one Head occurs. \( \mathrm{C} \) is the event at least two Heads occur \( \boldsymbol{P}(\boldsymbol{A})+\boldsymbol{P}(\boldsymbol{B})+\boldsymbol{P}(\boldsymbol{C})>1 \) Reason \( A, B \) and \( C \) are not Mutually Exclusive since events \( A \) and \( C \) have outcomes in common.
- A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
- B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
- C. Assertion is correct but Reason is incorrect
- D. Both Assertion and Reason are incorrect
Step-by-step solution
We compute P(A)=2/16=1/8, P(B)=4/16=1/4, P(C)=11/16. Sum=17/16>1, so assertion true. Events A and C share outcome HHHH, so they are not mutually exclusive, making reason true and explaining why sum exceeds 1.
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