Question

If three coins are tossed simultaneously find the probibality of getting at least one head

Answer 1

Answer:

E3 = {HTT, THT, TTH} and, therefore, n(E3) = 3. Therefore, P(getting 1 head) = P(E3) = n(E3)/n(S) = 3/8.

Step-by-step explanation:

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Answer 2

Answer:

7/8

Step-by-step explanation:

So, there are 8 possible situations (2 combinations of 3 coins)

Out of this, there is only 1 possible situation that all the coins will be head facing down. So 7/8th of the time, you should get at least 1 head.