 Hi, and welcome to the session. Today we will discuss the following question which says a pair of dice is thrown four times. If getting a doublet is considered a success, find the probability distribution of successes. Let's see its solution. We are given in the question that a pair of dice is thrown four times and getting a doublet is considered a success. So, let us assume that x denotes the number of doublets. When a pair of dice is thrown, then possible doublets are 1, 1, 2, 2, 3, 3, 4, 4, 5, 5 and 6, 6. Now the dice is thrown four times, so that means x can take the value 0, 1, 2, 3 and 4. Now the probability of getting a doublet will be 6 upon 36 because when a pair of dice is thrown, then the possible doublets are 6 and the possible outcomes are 36. So, this is equal to 1 upon 6. So, from this we have the probability of not getting a doublet will be 1 minus 1 upon 6 which is equal to 5 upon 6. Now, let us find probability when x is equal to 0 that is the probability of getting no doublet. This will be equal to 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 which is equal to 625 upon 1296. Next, let us find the probability when x is equal to 1 that is probability of getting 1 doublet, 3 non-doublets. This will be equal to 1 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 1 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 5 by 6 into 6 into 5 by 6 into 1 by 6 which is equal to 4 into 5 by 6 into 5 by 6 into 1 by 6 that is 4 into 125 upon 1296 now let us cancel 4 and 1296 by the common factor 4. So we get 125 upon 324 now we will find probability when x is equal to 2 that is the probability of getting 2 doublets and 2 non-doublets which will be equal to 1 by 6 into 1 by 6 into 5 by 6 into 5 by 6 plus 1 by 6 into 5 by 6 into 1 by 6 into 5 by 6 plus 1 by 6 into 5 by 6 into 5 by 6 into 1 by 6 plus 5 by 6 into 1 by 6 into 1 by 6 into 5 by 6 plus 5 by 6 into 1 by 6 into 5 by 6 into 1 by 6 plus 5 by 6 into 5 by 6 into 1 by 6 into 1 by 6 and this will be equal to 6 into 25 upon 1296. So here 6 and 1296 will get cancelled by the common factor 6. So we get 25 upon 216. Now we need to find out the probability when x is equal to 3 that is the probability of getting 3 doublets and 1 non-doublet which will be equal to 1 by 6 into 1 by 6 into 1 by 6 into 5 by 6 plus 1 by 6 into 1 by 6 into 5 by 6 into 1 by 6 plus 1 by 6 into 5 by 6 into 1 by 6 into 1 by 6 plus 5 by 6 into 1 by 6 into 1 by 6 into 1 by 6 which will be equal to 4 into 5 upon 1296. So let's cancel 4 and 1296 by the common factor 4. So we get 5 upon 324. Lastly let us find out the probability when x is equal to 4 that is the probability of getting 4 doublets and this will be equal to 1 by 6 into 1 by 6 into 1 by 6 into 1 by 6 which will be equal to 1 upon 1296. Thus the required probability distribution when x is equal to 0 probability is equal to 625 upon 1296 when x is equal to 1 the probability is 125 upon 324 when x is equal to 2 the probability is 25 upon 216 when x is equal to 3 the probability is 5 upon 324 and when x is equal to 4 the probability is 1 upon 1296. So this is the required answer to this question. With this we finished this session. Hope you must have enjoyed it. Goodbye, take care and have a nice day.