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 $\sum _3^{6}p(r)=p(3)+p(4)+p(5)+p(6)$
$=\left({}^6{C}_3{+}^6{C}_4{+}^6{C}_5{+}^6{C}_6\right)\frac{1}{2^6}=\left(2^6{-}^6{C}_0{-}^6{C}_1{-}^6{C}_2\right)\frac{1}{2^6}$
$=\left(64-1-6-15\right)\frac{1}{2^6}=\frac{42}{64}=\frac{21}{32}$
$\therefore f{\left(r\right)}_{r\ge 3}=N$$\sum _3^{6}P(B)=64$,$\frac{21}{32}=42$