Question

Eight coins are tossed at a time, the probability of getting at least 6 heads up, is

• A. $\frac{7}{64}$
• B. $\frac{57}{64}$
• C. $\frac{37}{256}$
• D. $\frac{229}{256}$

Answer 1

Answer: (C)

$\frac{37}{256}$

$x\le 6P(x=6)+P(x=7)+P(x=8)={{}^{8}C}_6(\frac{1}{2}{)}^8+{{}^{8}C}_7(\frac{1}{2}{)}^8+{{}^{8}C}_8(\frac{1}{2}{)}^8=(\frac{1}{2}{)}^8[28+8+1]=\left(37/256\right)$