Blackjack, or twenty-one as it is frequently called, is a popular gambling game played in Las Vegas casinos. A player is dealt two cards. Face cards (jacks, queens, and kings) and tens have a point value of 10. Aces have a point value of 1 or 11. A 52-card deck contains 16 cards with a point value of 10 (jacks, queens, kings, and tens) and four aces.

a. What is the probability that both cards dealt are aces or 10-point cards (to 4 decimals)?

b. What is the probability that both of the cards are aces (to 4 decimals)?

c. What is the probability that both of the cards have a point value of 10 (to 4 decimals)?

d. A blackjack is a 10-point card and an ace for a value of 21. Use your answers to parts (a), (b), and (c) to determine the probability that a player is dealt blackjack (to 4 decimals). (Hint: Part (d) is not a hypergeometric problem. Develop your own logical relationship as to how the hypergeometric probabilities from parts (a), (b), and (c) can be combined to answer this question.)

a) Since there are 52 cards, there are [tex] {52 \choose 2} = 1326 [/tex] ways to deal them; also, there are 16+4 = 20 cards that are aces or 10-point cards, thus, there are [tex] {20 \choose 2} = 190 [/tex] ways of dealing 2 cards that are 10-point cards or Aces. Since the probability for each individual hand is the same, then, the porbability that both cards dealt are 10-point cards or aces is 190/1326 = 0.1433.

b) There are [tex] {4 \choose 2} = 6 [/tex] ways to deal 2 aces. As a result, the probability that 2 aces are dealt is 6/1326 = 0.0045.

c) There are [tex] {16 \choose 2} = 120 [/tex] ways to deal two 10-point cards. Therefore, the probability that the two cards dealt are 10-point cards is 120/1326 = 0.0905.

d) There are 3 distinct ways of obtaining a pair of 10-point cards/Aces. Either you obtain two Aces, or you obtain two 10-point cards or you obtain 1 from each (which results in a balck jack). Since each possibility is mutually exclusive from the others, then

P(obtain 2 cards that are Aces or 10-point cards) = P( two 10-point cards) + P(two Aces) + P(Blackjack)

The probability to obtain 2 cards that are Aces or 10-point cards were computed in item (a) and it was 0.1433, and the other 2 probabilities besides blackjack were oobtained earlier in items (b) and (c). As a consequence

0.1433 = 0.0905 + 0.0045 + P(Blackjack)

So we conclude that

P(Blackjack) = 0.1433-0.0905-0.0045 = 0.0483

jmonterrozar jmonterrozar · Answer 2 · 2020-04-03T03:45:41+02:00

Answer:

a, 0.1433

b. 0.0045

c. 0.0905

d. 0.0483

Step-by-step explanation:

We have that the formula for hypergeometric probability is:

(r) (N - r)

(x) (n - x)

f (x) = ------------------

(N)

(n)

where r is "the number of individuals that y x value that the variable takes.

We have that the population size is 52. N = 52

The number of draws is equal to 2. n = 2

Now, knowing this we solve each one:

a. In this case the values are as follows: r = 20; x = 2

(20) (52-20)

(2) (2 - 2)

f (x) = ------------------ = 95/663 = 0.1433

(52)

( 2 )

b. In this case the values are as follows: r = 4; x = 2

(4) (52-4)

(2) (2 - 2)

f (x) = ------------------ = 1/221 = 0.0045

(52)

( 2 )

c. In this case the values are as follows: r = 16; x = 2

(16) (52-16)

(2) (2 - 2)

f (x) = ------------------ = 20/221 = 0.0905

(52)

( 2 )

d. The probability of blackjack would come being in relation to a, b, c

That is to say:

p (blackjack) = a - b - c

Replacing:

p (blackjack) = 0.1433 - 0.0045 - 0.0905 = 0.0483