There are three coins. One is a two-headed coin (having a 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.

Let P(A) be the probability of choosing a two-headed coin, P(B) be the probability of choosing biased coin that comes up with head 75% of times and P(C) be the probability of biased coin that comes up with tails 40% of the times

There are a total of 3 coins

⇒ P(A) = P(B) = P(C) = …(a)

Let H be the event of getting head after the coin toss

We have to find P(A|H) which means the probability of selecting two-headed coin given that we got heads after selecting and tossing the coin

By Baye's theorem

Now we have to find P(H|A), P(H|B) and P(H|C)

P(H|A) means getting heads given that coin A is chosen

Now coin A has both faces as head hence it will always show up heads

Hence P(H|A) = 1 …(b)

P(H|B) means getting heads given that coin B is chosen

Now it is given that coin B shows heads 75% of the times

Hence P(H|B) = 0.75

⇒ P(H|B) = 3/4 …(c)

P(H|C) means getting heads given that coin C is chosen

Now it is given that coin C shows tails 40% of the time

As there are only two options heads or tails on a coin hence remaining 60% will be for heads

Hence P(H|C) = 0.6

⇒ P(H|C) = …(d)

Substitute values from equation (a), (b), (c) and (d) in (i)

Hence the probability that the coin was chosen is two-headed is .

OR

First 6 positive integers are 1, 2, 3, 4, 5 and 6

As we have to select 2 numbers the number of ways of selecting two numbers is

Using

Hence the total number of ways is 15

Now X can take values 2, 3, 4, 5 and 6 (why not 1? Because there isn’t any number which is smaller than 1 in first positive integers)

Let X = 2

Which means that among the selected pair 2 is greater hence only 1 pair possible (1, 2)

X = 3

Which means that among the selected pair 3 is greater hence pair possible are (1, 3) and (2, 3)

X = 4

Which means that among the selected pair 4 is greater hence pair possible are (1, 4), (2, 4), (3, 4)

X = 5

Which means that among the selected pair 5 is greater hence pair possible are (1, 5), (2, 5), (3, 5) and (4, 5)

X = 6

Which means that among the selected pair 6 is greater hence pair possible are (1, 6), (2, 6), (3, 6), (4, 6) and (5, 6)

Representing it in a table

Mean is given by