00:01
Hi there, in this problem it is given that for a biased coin, head is three times likely to occur as tail and we have to find the expected number of tails if the coin is tossed twice.
00:10
So let us start with the solution to this problem.
00:14
Firstly, we have probability of getting a head is three times the probability of getting a tail and we know that probability of getting a tail and the probability of getting a head if the sum is taken it is equal to 1.
00:34
From here we will get 3 times probability of tail plus probability of tail is equal to 1 or probability of tail is equal to 1 divided by 4.
00:48
Then probability of getting a head will be 3 times the probability of getting a tail that is 3 divided by 4.
00:56
Let us introduce a random variable x which represents the number of tails occurred in 2 tosses.
01:05
So for two tosses we have, x can take the values 0, 1 or 2, that is we get no tails, 1 or 2 tails.
01:20
Now the probability that x is equal to 0 will be.
01:23
In the case when both the tosses come with a head, so we multiply 3 divided by 6 and 3 divided by 6 and we will get 9 divided by 16.
01:35
The probability of getting x is equal to 1 can be in the case when the first 1.
01:41
The first toss is head and the second is tail...