Question

6 coins are tossed together 64 times. If throwing a head is considered as a success then the expected frequency of at least 3 successes is

• A. 64
• B. 21
• C. 32
• D. 42

Answer 1

Answer: (D)

Here $p=\frac{1}{2}. q=1-q=1-\frac{1}{2}=\frac{1}{2}$
$n = 6, N = 64.$
Then $p\left(r\right)=^{n}C_{r}\,p^{r}\,q^{n-r}=^{6}C_{r}\left(\frac{1}{2}\right)^{r}\left(\frac{1}{2}\right)^{6-r}=^{6}C_{r}\left(\frac{1}{2}\right)^{6}$
$\therefore f \left(r\right)-Np\left(r\right)=64. ^{6}C_{r}. \frac{1}{64}=^{6}C_{r}$
Now $\displaystyle \sum_{3}^6 p(r)=p(3)+p(4)+p(5)+p(6)$
$=\left(^{6}C_{3}+^{6}C_{4}+^{6}C_{5}+^{6}C_{6}\right) \frac{1}{2^{6}}=\left(2^{6}-^{6}C_{0}-^{6}C_{1}-^{6}C_{2}\right) \frac{1}{2^{6}}$
$=\left(64-1-6-15\right) \frac{1}{2^{6}}=\frac{42}{64}=\frac{21}{32}$
$\therefore f \left(r\right)_{r\ge3}=N$$\displaystyle \sum_{3}^6P(B)=64$,$\frac{21}{32}=42$