Answer
$P( x \leq 3)$ = 0.9144
Work Step by Step
The probability of obtaining $x$ successes ${n}\choose{x}$ $\cdot$ $p^xq^{n-x}$
$P( x \leq 3)$ = P(0) + P(1) + P(2) + P(3)
P(0) = ${9}\choose{0}$ $\cdot$ $0.2^00.8^{9-0}$ = 0.1342
P(1) = ${9}\choose{1}$ $\cdot$ $0.2^10.8^{9-1}$ = 0.3020
P(2) = ${9}\choose{2}$ $\cdot$ $0.2^20.8^{9-2}$ = 0.3020
P(3) = ${9}\choose{3}$ $\cdot$ $0.2^30.8^{9-3}$ = 0.1762
Therefore, $P( x \leq 3)$ = 0.1342 + 0.3020 + 0.3020 + 0.1762 = 0.9144