Two are red cards and two are black cards

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Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these four cards are of the same suit, There are four suits i.e. Diamond, Spade, Heart, Club & 13 cards of each suit Since, they are different cases, So, we add the number of ways Thus, Required number of ways choosing four cards of same suit = 13C4 + 13C4 + 13C4 + 13C4 = 4 × 13C4 = 4 × 13!/(4!(13 − 4)) = 4 × 13!/(4! 9!) = 4 × (13 × 12 × 11 × 10 × 9!)/(4! × 3 × 2 × 1 × 9!) = 4 × (13 × 12 × 11 × 10)/(4 × 3 × 2 × 1) = 2860 ways Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these (ii) four cards belong to four different suits, Since, they are the same case, So, we multiply the number of ways Hence, Required number of ways choosing four cards from each suit = 13C1 × 13C1 × 13C1 × 13C1 = (13C1)4 = (13!/1!(13 − 1)!)^4 = (13!/1!12!)^4 = ((13 × 12!)/12!)^4 = (13)4 = 13 × 13 × 13 × 13 = 28561 ways Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these (iii) are face cards, King Queen and Jack are face cards Number of face cards in One suit = 3 Total number of face cards = Number of face cards in 4 suits = 4 × 3 = 12 Hence, n = 12 Number of card to be selected = 4 So, r = 4 Required no of ways choosing face cards = 12C4 = 12!/4!(12 − 4)! = 12!/(4! 8!) = (12 × 11 × 10 × 9 × 8!)/(4 × 3 × 2 × 1 × 8!) = (12 × 11 × 10 × 9 )/(4 × 3 × 2 × 1 ) = 495 ways Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these (iv) two are red cards and two are black cards, Since, they are the same case, So, we multiply the number of ways Total number of ways choosing 2 red & 2 black cards = 26C2 × 26C2 = (26C2)2 = (26!/(2! (26 − 2)!))^2 = (26!/(2! 24!))^2 = ((26 × 25 × 24!)/(2 × 1 × 24!))^2 = ((26 × 25)/(2 × 1))^2 = (13 × 25)2 = (325)2 = 105625 Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these (v) cards are of the same color? Since, choosing red OR black , they are different cases, So, we add the number of ways = 2 × 26!/4!(26 − 4)! = 2 × 26!/(4! 22!) = 2 × (26 × 25 × 24 × 23 × 22!)/(4 × 3 × 2 × 1 × 22!) = 2 × (26 × 25 × 24 × 23)/(4 × 3 × 2 × 1) = 29900

First, I do not fully understand the j guesses part. I assume it means that you play n rounds of the game, and must find the probability of 0, 1, 2, 3, 4 of the n games being won...?

You point to two cards, guessing which two are red and the other two black, and $j$ of those guesses actually are right.

$$\begin{array}{c}\blacksquare & \blacksquare & \blacksquare & \blacksquare \\ & \uparrow & \uparrow &\end{array}$$

As all configurations are equally likely, this reads as an ordered problem without replacement when picking cards in each round. Therefore probability of selecting two red cards is 1/6

Yes. That is the probability that $j=4$. You pointed to both red cards and did not point at both black cards.

$$\mathsf P(j{=}4) = \frac 16$$

$$\begin{array}{c}\clubsuit & \color{red}\diamondsuit & \color{red}\heartsuit & \spadesuit \\ & \uparrow & \uparrow &\end{array}$$

I think the general answer is P(j wins of n Games) = (1/6)j * (5/6)(n-j) but I have invented the variable n because the initial question appears vague.

No, you are looking for the probabilities of guessing the colour of 0,1,2,3,or 4 cards by pointing to the two red cards. You have done the latter. Find the others.

$$\mathsf P(j{=}0) = \\ \mathsf P(j{=}1) = \\ \mathsf P(j{=}2) = \\ \mathsf P(j{=}3) = $$