 Hello and how are you all today? My name is Priyanka. The question to be discussed is 1500 families with two children were selected randomly and the following data were recorded. Now here the number of girls in a family are 2, 1, 0 respectively and the number of families in 4, 75, 8, 1, 4 and 2 double 1. What we need to do is we need to compute the probability of a family chosen at random having 2 girls, 1 girl and no girl. So before proceeding on with the solution we should be well versed with the key idea that is the formula for finding the probability of agreeing an event that is 2 girls in the first case, 1 girl in the second case and no girl in the third case. The formula is number of possible outcomes divided by total number of outcomes. Right? So this is the key idea to this question. Let us proceed on with our solution. Now here first of all we need to find out for 2 girls. The number of families having 2 girls are 4, 75 out of the total families. So first of all we will be finding out the total number of families. So the total number of families are 4, 75 plus 8, 1, 4 plus 2 double 1 that is further equal to 1500. So out of 1500 families 4, 75 are having 2 girls. So the first points answer that is the probability of having 2 girls is 4, 75 families out of the total 1500 that is further equal to 19 upon 60 that is on simplifying it out. Right? Second point is to find out the probability of 1 girl, a family that is having 1 girl. So it is 8, 1, 4 divided by 1500 that is further equal to on simplifying 4, 0, 7 divided by 750. And lastly we need to find out the probability that the family does not have any girl. So there are 2 double 1 families like that out of the 1500 and they cannot be, they are not having any common factor so the answer will remain the same. So the final answer is for the first part 19 upon 60, for the second part it is 4, 0, 7 upon 750 and for the third part it is 2 double 1 upon 1500. This is the required answer for this session. This completes. Hope you have a great day ahead and take care.