00:01
In this problem, it is said that a coin is biased in such a way that a head is twice as likely to occur as a tail.
00:08
We need to find the expected number of tails when this coin is twice.
00:12
Now, first of all, if this coin has twice the probability of obtaining a head as the probability of obtaining a tail, then that will mean that the probability of obtaining a head will be two times the probability of obtaining a tail.
00:28
And since the probability of obtaining a head and the probability of obtaining a tails should the sum of these should be equal to one because when you toss a coin you will either get a head or a tail so the sum of these two probabilities should be one probability of heads is 2 pt so 2 pt plus pt is equal to 1 that implies that 3 pt is equal to 1 which means that the probability of obtaining a tails is is 1 divided by 3 and thus the probability of getting our heads will be 2 pt which is 2 times 1 by 3 which is 2 divided by 3 now let us consider x to be the random variable of obtaining a tail so the probability of x equals to 0 means the probability of getting 0 p's so that means we will have a head on the first throw and a head on the second toss of the coin so that will be 2 by 3 times 2 by 3 because the probability of obtaining ahead on the first toss is 2 by 3.
01:36
The probability on obtaining ahead on the second toss is 2 by 3.
01:39
So if we take both of these events together, we use the multiplication rule to multiply them and we will get 4 by 9.
01:47
Next, the probability of x equals to 1 is the probability of getting 1 tail.
01:52
So one possibility is getting ahead on the first coin and the tail of the tail of the tail.
01:59
On the second toss.
02:00
So the probability will be 2 by 3 times 1 by 3.
02:03
And there is another option which is a tail on the first toss and a head on the second toss.
02:09
So we add the probabilities of these two options in order to get the total probability...