Answer:
a) The porbability that both cards dealt are 10-point cards or aces is 0.1433
b) The probability that 2 aces are dealt is 0.0045
c) The probability that the two cards dealt are 10-point cards is 0.0905
d) The probability that a player is dealt a blackjack is 0.0483
Step-by-step explanation:
a) Since there are 52 cards, there are ways to deal them; also, there are 16+4 = 20 cards that are aces or 10-point cards, thus, there are
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 ways to deal 2 aces. As a result, the probability that 2 aces are dealt is 6/1326 = 0.0045.
c) There are 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