A coin is tossed twice. Events E and F are defined as follows : E = heads on first toss, F = heads on second toss. Find the probability of E∪F.

Engineering

Mathematics

Conditional Probability

Question

Hence p $(E \cup F) = \frac{1}{4}$

Hence p $(E \cup F) = \frac{3}{4}$

Hence p $(E \cup F) = \frac{1}{2}$

none of these

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Let H denotes head and T denotes tail.
Sample space S = {(H,H), (H,T), (T,H), (T,T)}
Total number of sample points n(S) = 4
Now, E = {(H,H), (H,T)}
and F = {(H,H), (T,H)}
E∪F = {(H,H), (H,T), (T,H)}
So n(E∪F) = 3
Hence $P (E \cup F) = \frac{n (E \cup F)}{n (S)} = \frac{3}{4}$

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