Question

A man tosses a coin 10 times, scoring 1 point for each head and 2 points for each tail. Let P(k) be the probability of scoring at least k points. What is the largest value of K such that $P(k) > \dfrac{1}{2}$?
(a). 14
(b). 15
(c). 16
(d). 17

Answer 1

Hint: Find the number of ways the score is greater than k and then find the total number of ways of throwing 10 coins. Then, find the largest value of k such that $P(k) > \dfrac{1}{2}$.

Complete step-by-step answer:
We know that the total number of ways of tossing a coin 10 times is ${2^{10}}$, which is 1024.
$N(S) = 1024............(1)$
It is given that 1 point is given for one head and 2 points for a tail. Hence, one toss can be represented by $x + {x^2}$ having two possibilities. Then, 10 tosses are given as follows:
10 tosses = ${(x + {x^2})^{10}}$
The number of possibilities with value k is given as the coefficient of ${x^k}$ in the expression ${(x + {x^2})^{10}}$.
${(x + {x^2})^{10}}$ can also be written as ${x^{10}}{(1 + x)^{10}}$. Hence, the number of possibilities of getting the value K is the coefficient of ${x^{k - 10}}$ in the expression ${(1 + x)^{10}}$.
Using the binomial theorem, we know that the coefficient of ${x^{k - 10}}$ in the expression ${(1 + x)^{10}}$ is given as ${}^{10}{C_{k - 10}}$.
Hence, the number of possibilities of the score being at least k is given as follows:
$N(k) = {}^{10}{C_{k - 10}} + {}^{10}{C_{k - 9}} + ....... + {}^{10}{C_{10}}..........(2)$
The probability is given as the ratio of the number of favorable outcomes to the total number of outcomes.
$P(k) = \dfrac{{N(k)}}{{N(S)}}$
From equations (1) and (2), we have:
$P(k) = \dfrac{{{}^{10}{C_{k - 10}} + {}^{10}{C_{k - 9}} + ....... + {}^{10}{C_{10}}}}{{1024}}...........(3)$
Using the formula for the last half binomial coefficients for even n, we know that:
$\dfrac{{{}^n{C_{\dfrac{n}{2}}}}}{2} + {}^n{C_{\dfrac{n}{2} + 1}} + {}^n{C_{\dfrac{n}{2} + 2}} + .......{}^n{C_n} = {2^{n - 1}}$
Hence, for n = 10, we have:
$\dfrac{{{}^{10}{C_5}}}{2} + {}^{10}{C_6} + {}^{10}{C_7} + ....... + {}^{10}{C_{10}} = {2^{10 - 1}}$
$\dfrac{{{}^{10}{C_5}}}{2} + {}^{10}{C_6} + {}^{10}{C_7} + ....... + {}^{10}{C_{10}} = {2^9}$
By adding another term $\dfrac{{{}^{10}{C_5}}}{2}$, we obtain:
${}^{10}{C_5} + {}^{10}{C_6} + {}^{10}{C_7} + ....... + {}^{10}{C_{10}} > {2^9}$
Dividing both sides by ${2^{10}}$, we get:
$\dfrac{{{}^{10}{C_5} + {}^{10}{C_6} + {}^{10}{C_7} + ....... + {}^{10}{C_{10}}}}{{{2^{10}}}} > \dfrac{{{2^9}}}{{{2^{10}}}}$
Simplifying, we have:
$\dfrac{{{}^{10}{C_5} + {}^{10}{C_6} + {}^{10}{C_7} + ....... + {}^{10}{C_{10}}}}{{1024}} > \dfrac{1}{2}$
Comparing the above equation with equation (3), we have:
$k - 10 = 5$
Then, solving for k, we have:
$k = 10 + 5$
$k = 15$
Hence, the value of k is 15.
Hence, option (b) is the correct answer.

Note: For $P(k) > \dfrac{1}{2}$, the number of favorable outcomes should be greater than ${2^9}$. Hence, you can use this number to equate to the number of favorable outcomes and find k by this method also.