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Q)

# A die is thrown 10 times. If getting a prime number is considered a success, find the probability of getting not more than 8 successes.

$(A)\; \frac{1010}{1024}$
$(B)\; \frac{1013}{1024}$
$(C)\; \frac{1000}{1024}$
$(D)\; \frac{1015}{1024}$

Given :$n = 10 ,\; \; \; \; \; P = \frac{1}{2}, \; \; \; \; \; q = 1 - \frac{1}{2} = \frac{1}{2}$
$P(x \leq 8)$ = 1 - {$P (x = 9) + P (X = 10)$}
$= 1 -\{10C_9 (\frac{1}{2})^9 . \frac{1}{2} + 10C_{10} (\frac{1}{2})^{10} \}$
$= 1 - (\frac{1}{2})^{10} (11) = 1 - \frac{11}{1024}$
$= \frac{1013}{1024}$