 Hello and welcome to the session. The given question says A and B toss a coin alternately till one of them gets a head and wins the game. If A starts first, find the probability that B will win the game. Let's start with the solution and we know that when a coin is tossed, possible outcomes are head and tail. Let us denote the head by capital H and the tail by capital T. So the probability of getting a head is 1 by 2 and probability of getting a tail is also 1 by 2. Therefore probability of not getting a head is given by 1 minus probability of getting a head that is 1 minus half which is equal to half. Similarly probability of not getting a tail is 1 minus probability of tail which is again equal to half. Let E denotes that A gets a head, F denotes that B gets a head. Now A starts tossing the coin first so you will get the first chance then B tosses the coin and again the third chance A will toss the coin. So A will toss the coin in the first, third, fifth, seventh all these chances and B will toss the coin A that is in the second chance if in the first row A does not get a head then in the again fourth row if in the third row he does not get a head and in the second row also B does not get a head so on. Therefore E in the first toss is half. Now it fails in the second that is he gets a head that is probability of getting a head in the third row by A is given by probability of E complement intersection probability of F complement intersection E that is probability of not getting a head in the first row not getting a head in the second row by B and then probability of getting a head in the third row. So this is equal to probability of E complement into probability of F complement into probability of E. So this is equal to half into half into half which gives 1 by 2 cube. Similarly probability of getting a head in the fifth row which is by A such that a head do not appear in the first four throws is given by probability of E complement intersection F complement intersection E complement intersection F complement intersection E that is probability of not getting a head in the first, second, third and fourth row and probability of getting a head in the fifth row. So this is equal to probability of E complement into probability of F complement into probability of E complement into probability of F complement into probability of E and all these probabilities are equal to half so we have half raised to the power 5. Hence probability winning the game by A is equal to probability of E union probability of E complement intersection F complement intersection E union probability of E complement intersection F complement intersection E complement intersection intersection F complement intersection E union and so on and this is further equal to probability of E plus probability of E complement intersection F complement intersection E plus probability of E complement intersection F complement intersection E complement intersection F complement intersection A plus so on and this is further equal to half, half raise to the power 3 plus half raise to the power 5 and so on. Now this is a geometric progression, so solving it. This is equal to half divided by 1 minus half raise to the power 2 and this is equal to half divided by 1 minus 1 by 4 and this gives half divided by 3 by 4 which is equal to half into 4 by 3. So we have 2 by 3. So probability that A wins the game is 2 by 3. Thus probability, we have to find the probability that B wins the game. Therefore probability of B winning the game is equal to 1 minus probability of A winning the game and this is equal to 1 minus 2 by 3 which is equal to 1 by 3. Hence our answer is 1 divided by 3. So this completes the session by antique kill.