In the game of blackjack played with one deck, a player is initially dealt 2 diferent cards from the 52 different cards in the deck a winning ℡ack ack 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?
the probability is □ type an integer or a simplified fraction)
the percentage of hands that are winning blackjack hands is round to two decimal places as needed.)

4.83%

Step-by-step explanation:

Total number of possible hands

we are interested in selecting two of the 52 cards in the deck.

n=52

r=2

therefore,

52C_2 = 52!/(2!50!)= 1326 possible hands

therefore total number of blackjacks

A blackjack consists of 1 of 4 aces and 1 of 16 other cards that are worth 10 points.

by fundamental counting rule

4*16= 64 possible blackjacks

2)% of hands that are winning blackjack hands =(32/663)*100 =4.83%

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: