In the game of blackjack played with one deck, a player is initially dealt 2 different cards from the 52 different cards in the deck. a winning "blackjack" hand is won by getting 1 of the 4 aces and 1 of 16 other cards worth 10 points. the two cards can be in any order. find the probability of being dealt a blackjack hand. what approximate percentage of hands are winning blackjack hands?

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

Where n is the total number of letters and r are chosen from them

The number of ways to choose 1 As is:

The number of ways to choose a 10-point letter is:

Therefore, the number of ways to choose an Ace and a 10-point card is:

Now the number of ways to choose any 2 cards from a deck of 52 cards is:

Therefore, the probability of obtaining an "blackjack" is:

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 =

So 1 out of the 16 cards worth 10 points will be equal to =

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) = = 0.04827

Percentage of getting blackjack hand = 4.83%

i dont get the question

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6.25 grams

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