Question

A man tosses a coin $10$ times, scoring $1$ point for each head and $2$ points for each tail. Let $P\left(K\right)$ be the probability of scoring at least $K$ points. The largest value of K such that $P\left(K\right)>\frac{1}{2}$ is.

• A. 14
• B. 15
• C. 16
• D. 17

Answer 1

Answer: (A)

14

For sum $=k$

let tail occurs n times then head will occur $10-n$ times

Then $10-n+2n=k$

$k=10+n$

P(K) is the probability of scoring atleast K point , if we subtract all the cases in which all head occurs then we get the desired P(K) , only in the case in which all head occcurs our sum will be less than k

$P\left(K\right)=1-\left(C\right(10,0)\times {\left(\frac{1}{2}\right)}^0\times {\left(\frac{1}{2}\right)}^{10}+C(10,1)\times {\left(\frac{1}{2}\right)}^1\times {\left(\frac{1}{2}\right)}^9..............C(10,n-1)\times {\left(\frac{1}{2}\right)}^{10})$

$P\left(K\right)=1-{\left(\frac{1}{2}\right)}^{10}\left(C\right(10,0)+C(10,1).............C(10,n-1\left)\right)$

and we have $n=k-10$

$P\left(K\right)=1-{\left(\frac{1}{2}\right)}^{10}\left(C\right(10,0)+C(10,1).............C(10,n-1\left)\right)$

$P\left(K\right)>\frac{1}{2}$

$1-{\left(\frac{1}{2}\right)}^{10}\left(C\right(10,0)+C(10,1).............C(10,k-11\left)\right)>\frac{1}{2}C(10,0)+C(10,1).............C(10,k-11)<2^9$

Now the best approach to find the value of k is to check by options

so by hit and trail method

$k=14$