Three unbiased coins are tosseed. The probability of getting at most two heads is …………

A) 3/8

B) 1/2

C) 3/4

D) 7/8

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2 Answers

Correct option is: D) \(\frac{7}{8}\)

\(\therefore\) Possible outcomes are (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT)

\(\therefore\) n (s) = Total No of outcomes = 8.

Let E be the event of getting at most two heads 

E = (HHT, HTH, THH, HTT, THT, TTH, TTT)

\(\therefore\) n (E) = 7

\(\therefore\) P (E) = \(\frac {n(E)}{n(S)} = \frac 78\) 

Hence, the probability of getting at most two heads is \(\frac{7}{8}\)

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Correct option is: D) \(\frac{7}{8}\)

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