 Good morning friends I am Pooja and today we will work out the following question. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale otherwise it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and three are bad ones will be approved for sale. Now the multiplication rule for three events E, F and G is given by probability of E intersection F, intersection G is equal to probability of E into probability of F given E into probability of G given E intersection F. So this is the key idea behind our question. Let us begin with the solution now. Now let A be the event of selecting first good orange, B be the event of selecting second good orange and C be the event of selecting third good orange. Now we are given that total number of oranges is 15 out of which 12 are good and 3 are bad ones. We are also given that if all the three oranges are good the box is approved for sale otherwise it is rejected. And we have to find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale. Now probability of A that is probability of selecting first good orange is equal to 12 upon 15. Because we are given that total number of oranges is 15 out of which 12 are good ones so probability of selecting first good orange out of these 15 oranges will be 12 upon 15. Now we are given that second orange is drawn without replacement so we have to find probability of B given that A has occurred that is to find probability of B given A. Now B is the event of selecting second good orange and we are given that the second orange is drawn without replacement. So we have to find the probability of selecting second good orange given that the first good orange is already selected. That is we have to find probability of B given A. Now probability of B given A is equal to 11 upon 14. Now since the second orange is selected without replacement so the total number of oranges left out in the box will be 14 out of which number of good oranges will be 11. Because out of 12 good oranges one good orange has already been selected so we get probability of B given that A has already occurred is equal to 11 upon 14. Now again the third orange is drawn without replacement so we have to find probability of C given that A and B have already occurred that is to find probability of C given A intersection B. Now C is the event of selecting third good orange and we are given that this third good orange is selected without replacement. So we have to find the probability of selecting third good orange given that the first and second good oranges have already been selected. That is we have to find probability of C given A intersection B. Now this probability of C given A intersection B is equal to 10 upon 13. Now since this third orange is selected without replacement so the total number of left out oranges in the box will be 13. And since out of 12 good oranges two good oranges have already been selected so the total number of good oranges will be 10. So the probability of selecting this third good orange given that the first and second good oranges have already been selected is equal to 10 upon 13. Now by key idea we know that the multiplication rule for three events E, F and G is given by probability of E intersection F intersection G is equal to probability of E into probability of F given that E has already occurred into probability of G given that E and F have already occurred. So here probability of selecting all the three good oranges without replacement will be given by probability of A intersection B intersection C is equal to probability of A into probability of B given that A has occurred into probability of C given that A and B have already occurred. Now this is equal to probability of A is equal to 12 upon 15 so we have 12 upon 15 into probability of B given A is equal to 11 upon 14 so we have 11 upon 14 into probability of C given A and B have already occurred is equal to 10 upon 13 so we get 10 upon 13. And we get this is equal to 44 upon 91 so we have got our answer as 44 upon 91 hope you have understood the solution by and take care.