 Hello and welcome to the session. In this session, we will discuss a question which says that in an experiment of wearing two undies dies, there is the probability that total has nine appears undies given that four appears on one of the dies. Now before starting the solution of this question, we should know a result. And that is conditional probability P of occurrence of event A given that event B occurs is equal to probability P of event A intersection B upon probability P of event P. Now this result will work out as a key idea for solving out the given question. Now let us start with the solution of the given question. Now the question is given that two dies are weird simultaneously we have to find the probability that total of nine appears and four appears and one of the dies. Now let us write one point state for the given question. Now we know that whenever we roll a die the total number of outcomes is now when two dies are rolled simultaneously then the possible outcomes are getting a number one on first die and getting a number one on second die then getting a number one on first die and getting a number two on second die then getting a number one on first die and getting a number three on second die. So here we have written all possible outcomes in rolling of two dies. Now here the sample space will contain all these possible outcomes that is if a set containing all these ordered pairs and there are 36 ordered pairs. So total number of outcomes is equal to 36. Now let us define the events. Event P is total of nine appears on the dies and event B is number four appears. One of the dies to fall let us write the elements for event A. For this we have to check those ordered pairs in the sample space S whose sum of first and second components is equal to nine. And here you can see three plus six is nine, four plus five is nine, then five plus four is also nine and six plus three is nine. So event A is a set containing all the pairs, three six, four five, five four and six three. Now let us write elements of event B. Now this is event B. Now for writing elements of event B we have to check those ordered pairs of sample space S whose one of the components is four. So event B is a set containing all the pairs one four, two four, three four, four one, four two, four three, four four, four five, four six, five four, six four. Now let us write event A intersection B which is a set containing ordered pairs that are common to events A and B. Those ordered pairs are four five five four. So event A intersection B is a set containing ordered pairs four five and five four. Now we know that the total number of outcomes is equal to thirty six and number of outcomes is equal to event B is equal to eleven. So probability of event B is equal to number of outcomes is equal to event B that is eleven upon total number of outcomes that is thirty six. So probability of event B is equal to eleven upon thirty six. Now number of outcomes is equal to event A intersection B is equal to probability P of event A intersection B is equal to number of outcomes is equal to event A intersection B that is two upon total number of outcomes that is thirty six. So probability of event A intersection B is equal to two upon thirty six. Now we have defined event A as total of nine appears on the dice and event B as four appears on one of the dice and we have defined the probability that total of nine appears on dice. Given that four appears on one of the dice it means we have to find conditional probability of occurrence of event A. Given that event B occurs and using this result which is given to us in key idea we know that this is equal to probability of event A intersection B upon probability of event B. Now this is probability of event B and this is probability of event A intersection B. So conditional probability of occurrence of event B occurs is equal to probability of event A intersection B upon probability of event B. This is equal to two upon thirty six four upon eleven upon thirty six and on solving this is equal to two upon eleven. So conditional probability of occurrence of event B occurs is equal to two upon eleven. Now this is the required answer that is probability that total of nine appears on the dice given that four appears on one of the dice is equal to two upon eleven. Now here you can see there are eleven elements in event B and conditional probability is two eleventh of B's outcomes and that particular two outcomes are the ordered pairs four five and five four which are also in event A. This is the solution of the given question and that's all for this session hope you all have enjoyed the session. Thank you.