In this problem, we are going to determine the probabilities of certain events. Now in this problem, cars are built one at a time from a standard 52 card deck. And the first problem, we need to determine the probability that if the first two cards are both space, then what, what is the probability that the next three cards are also spades? So in order to find the probability first of all, we need to find the number of favorable outcomes. Now in a standard 52 card deck, there are 13 speeds. And since the first two cards are both space, that means out of these 13 spades, the number of spades left is 13 -2, which is left. And since we need to find the probability that the next three cards of space, that means we need to choose three speeds out of the remaining 11 speeds. So the number of ways that can be done is equals to 11 C3. Now the total number of outcomes also needs to be determined. There are 52 cards and two cars have already been drawn, so that means that there are 50 cards remaining and out of that three cars need to be selected. So that means the total number of outcomes will be 50 C3. So if this is calculated then 11 C3 is equals to 165 and 50 C3 is equals to 196 00 So 1 65 divided by 19600 is approximately equals to 0.84 So that is the required probability in this case. Now, in the second problem, If the fullest three cards of space and what is the probability that the next two cards are also speeds? So we first of all need to find the number of favorable outcomes. Out of 13 spades, three spades have already been selected, so the remaining number of speeds is 13 -3, close to 10. And out of that we need to find the next two cards are speed. So out of those 10 remaining speeds, we need to select to space, so a number of ways that can be done is equal to 10 C three and the total number of outcomes since there are 52 cards in the deck and three of them have already been selected. So the remaining number of cards is 50 to minus three, which is equal to 49. And out of that we have to choose a total or two cards, so that's 49 C. To actually the numerator also needs to be that tendency to we need to choose two cards out of the remaining 10 spades and we need to choose two cars are the remaining 49 cards, So the value of 10 C. Two is equals to 45 and the value of 40 92 is equal to 1176. And the value of 451176 is approximately, it goes to 0.0383. So that is the probability in this case. And in the last problem, four cars are selected, They are all spades. What is the probability that the next card is also a spade? So in this case four cards and spades are 13 cars. Four of them have already been selected out of 13 spades, four have been selected, So the remaining number of spades will be 30 minus four, which is nine, so that's nine C. One because we only need to choose one more car. And Out of 52 cards, four cars have been drawn, so there are 48 cards in the deck, and out of that we have to choose one. So 48 C1 is the total number of outcomes, So 91 is nine and 48 C1 is 48 and nine, divided by 48 is equals to 0.1875. So that is the probability in this case.