There are three c

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