Two coins are simultaneously tossed until one of them comes...

Two coins are simultaneously tossed until one of them comes up a head and the other a tail. The first coin comes up a head with probability p and the second with probability q. All tosses are assumed independent. (a) Find the PMF, the expected value, and the variance of the number of tosses. (b) What is the probability that the last toss of the first coin is a head?

Transcript

00:01 So the approach that i'll take here is to begin by considering what our sample space is.

00:06 So we should know we have four different outcomes.

00:09 Could get tails, tails, heads, tails, or hails, pardon me, tails, tails, heads, or heads, heads.

00:22 Now i should note that each one of these outcomes is not necessarily, it is not necessarily equal in probability.

00:32 So particularly, we should know that for getting tails on the first coin, p of t1, that is equal to 1 minus p, and getting tails on the second coin is equal to 1 minus q, where those values are not the same.

00:50 That being said, if we're looking for the probability of getting heads, tails, or tails heads, each one of those events would be, well, for heads tails, probability of h1t2, that would be equal to, well, probability of getting heads on the first is just p, and tails on the second would be one minus q, and probability of tails on the first and heads on the second would be equal to one minus p times q.

01:24 So we have overall the probability of our desired event, which i'll just call event e, is going to be equal to p times 1 minus q plus 1 minus p times q.

01:37 So if we expand that out, we'd get p minus p q plus q minus q.

01:47 So we can simplify this down as p minus 2pq plus q.

01:54 That being said, now that we know the probability of our desired event, occurring, we can now think of this as being a geometric distribution with probability of success.

02:10 I'll call this p prime.

02:12 That's the probability of success on the geometric distribution being, pardon me, i'll put it this way.

02:20 This would now be a geometric distribution with the probability that the number of trials until our desired outcome occurs being equal to, so we should have it's one or actually it should be x equals little k just for consistency that would be equal to one minus the probability of success so one minus p plus two pq oops two p q minus q minus q to the power of k minus one times the probability of success so p minus two p q plus q so in the that case, we would also have that the expected value of the number of trials until a success is going to be one over the probability of success.

03:14 That's one over p minus 2 pq plus q...

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