Suppose the two cards dealt to you are of different denominations. what is the probability that you'll be able to make four of a kind out of all seven cards available to you?

The probability of getting a four of a kind is 29/23030 = 0.00126.  

Step-by-step explanation:

The poker deck has 52 cards, divided in 4 suits, and each suit has 13 different denominations.

In this exercise you are given that the 2 cards you are dealt have different denominations, lets call them A and B, so in order to make a 4 of a kind, you need to get from the 5 remaining community cards, 3 cards that have the denomination of one of A or B.

You cant get a four of a kind with two different denominations at the same time, because you need to reserve the majority of the community cards for just one of them. So each four of a kind is a disjoint event from the others.

Recall that, apart from the obvius four of a kind that you can get with A and B, you can still get four of a kind with any other denomination, because 4 of the community cards can still have the same denomitation.

Lets first calculate the probability to get a four of a kind with A. With B it is the same computation and the result will, therefore, be the same.

First, note that the order in which the cards are given doesnt matter, so we can calculate the probability of getting triple A in the flop (in other words, the first 3 given cards), and then multiply by all possible permutations.

Since you start with one A in your hand, there are 3 in the deck of 50 cards (remove the two cards given to you). The probability of drawing the first A is 3/50. The probability to draw the second A once you drew the first one is 2/49 (because you removed another A from the deck). The probability for the third A to be drawn is 1/48. This gives us a probability of

3/50 * 2/49 * 1/48 = 1/19600

of drawing the 3 A in the first 3 given cards. The other 2 cards can be anything.

To draw 3 aces in 5 cards we need to multiply be the total amount of ways to put 3 cards in a set of 5. That number is equivalent to the number of ways to pick places for the three A ignoring the order for them. That is, So, in total, the probability to get a four of the kind A is

10 * 1/19600 = 1/1960

Similarly, the probability to get a four of a kind with B is 1/1960.

To get a four of a kind with any other kind, we need first to specify the kind; we have 13-2 = 11 possibilities to choose (we substract the 2 kinds A and B). Then we need to specify how the cards of that kind will be placed among the community cards. We have as many possibilities as the total amount of ways to pick where the other card will be, that is, 5 possibilities. Hence, we have 55 possibilities to make a four of a kind with a kind different than A or B.

We need to calculate now what is the probability of getting 4 cards of a specific kind in a specific way, for example, the first four cards. That probability is

4/50 * 3/49 * 2/48 * 1/47 = 1/230300

because we start with 50 cards on the deck with 4 cards of them being of the kind we need, and we remove them one by one.

As a result, the probability of getting a four of a kind with a different kind than A and B is 55/230300 = 11/46060.

By summing disjoint cases, We conclude that that the probability to get a four of a kind is

11/46060 + 1/1960 + 1/1960 = 29/23030 = 0.00126  

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺

answer is b

step-by-step explanation: