Answer:
[tex]P =4.83\%[/tex]
Step-by-step explanation:
First we calculate the number of possible ways to select 2 cards an ace and a card of 10 points.
There are 4 ace in the deck
There are 16 cards of 10 points in the deck
To make this calculation we use the formula of combinations
[tex]nCr=\frac{n!}{r!(n-r)!}[/tex]
Where n is the total number of letters and r are chosen from them
The number of ways to choose 1 As is:
[tex]4C1 = 4[/tex]
The number of ways to choose a 10-point letter is:
[tex]16C1 = 16[/tex]
Therefore, the number of ways to choose an Ace and a 10-point card is:
[tex]4C1 * 16C1 = 4 * 16 = 64[/tex]
Now the number of ways to choose any 2 cards from a deck of 52 cards is:
[tex]52C2 =\frac{52!}{2!(52-2)!}[/tex]
[tex]52C2 = 1326[/tex]
Therefore, the probability of obtaining an "blackjack" is:
[tex]P = \frac{4C1 * 16C1}{52C2}[/tex]
[tex]P = \frac{64}{1326}[/tex]
[tex]P = \frac{32}{663}[/tex]
[tex]P =0.0483[/tex]
[tex]P =4.83\%[/tex]
Answer:
Probability = 0.0483
Percentage = 4.83%
Step-by-step explanation:
We know that a blackjack hand played with one deck consists of:
1 of the 4 aces = [tex]\frac{4}{52}[/tex]
So 1 out of the 16 cards worth 10 points will be equal to = [tex]\frac{16}{52}[/tex]
Finding the probability of getting a blackjack hand assuming that the cards were not replaced:
P (blackjack hand) = P(1st ace) × P(2nd 10 point card) + P(1st 10 point card) × P(2nd ace)
P (blackjack hand) = [tex]\frac{4}{52} \times \frac{16}{51} + \frac{16}{52} \times \frac{4}{51}[/tex] = 0.04827
Percentage of getting blackjack hand = 4.83%
