Question

A coin is tossed three times. The probability of getting atmost 3 heads

Answer 1

Answer:

Required Probability= 3c0(1/2)³ +

3C1(1/2)³ + 3C2(1/2)³ + 3C3(1/2)³

=

$\frac{1}{8} (1 + 3 + 3 + 1) = \frac{1}{8} \times 8 = 1 \: ans.$