Question

There are three coins. One is two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?

OR

Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained.
Find the probability distribution of the random variable X, and hence find the mean of the distribution.

Answer 1

Let $E_1$: choosing first (two headed) coin, $E_2$ : choosing second (biased) coin, $E_3$ : choosing third coin. Also, let A : the coin showing heads.
$\therefore P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3},P\left(A|E_1\right)=1,P\left(A|E_2\right)=\frac{75}{100},P\left(A|E_3\right)=\frac{60}{100}$.

By Bayes' Theorem, $P\left(E_1|a\right)=\frac{P\left(A|E_1\right)P\left(E_1\right)}{P\left(A|E_1\right)P\left(E_1\right)+P\left(A|E_2\right)\left(P\right(E_2)+P(A|E_3\left)P\right(E_3)}$
$=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{3}{5}}=\frac{20}{47}$.

OR

Total number of ways of selecting two numbers form six positive integers = ${}^6{C}_2=\frac{6\times 5}{2}=15$ .
Let X denotes the larger of the Two numbers selected, So, values of X:2,3,4,5,6.
So, $P(X=2)=\frac{1}{15},P(X=3)=\frac{2}{15},P(X=4)=\frac{3}{15},P(X=5)=\frac{4}{15},P(X=6)=\frac{5}{15}$
$\therefore$ distribution can be written as:
$\begin{array}{cccccc}& & & & & \\ X& 2& 3& 4& 5& 6\\ P\left(X\right)& \frac{1}{15}& \frac{2}{15}& \frac{3}{15}& \frac{4}{15}& \frac{5}{15}\end{array}$

Therefore, mean of the distribution =$\sum XP\left(X\right)=2\times \frac{1}{15}+3\times \frac{2}{15}+4\times \frac{3}{15}+5\times \frac{4}{15}+6\times \frac{5}{15}=\frac{70}{15}=\frac{14}{3}$