two cards are selected at random without replacement from a well-shuffled deck of 52 playing cards. find the probability of the given event. a. two cards of the same suit are drawn. b. a pair is not drawn.
To find the probability of the given events, let's consider each event separately:
a. Two cards of the same suit are drawn:
There are 4 suits in a standard deck of playing cards (hearts, diamonds, clubs, and spades). Each suit has 13 cards (Ace, 2-10, Jack, Queen, King).
To calculate the probability of drawing two cards of the same suit, we first need to determine the total number of possible outcomes (ways to draw two cards without replacement). For the first card, we can select any of the 52 cards. For the second card, after drawing the first card, there will be 51 cards remaining in the deck.
So, the total number of possible outcomes = 52 * 51 = 2,652.
Next, we need to determine the favorable outcomes (ways to draw two cards of the same suit).
- We can choose any of the 4 suits for the first card. Once the suit is selected, there are 13 cards of that suit remaining.
- For the second card, out of the remaining 12 cards in that suit, there is only one card that matches the suit of the first card.
So, the number of favorable outcomes = 4 * 13 * 1 = 52.
Therefore, the probability of drawing two cards of the same suit = favorable outcomes / total outcomes = 52 / 2,652 ≈ 0.0196 (or about 1.96%).
b. A pair is not drawn:
To find the probability of not drawing a pair, we need to calculate the favorable outcomes (ways to draw two cards that do not match in rank) and divide it by the total number of outcomes.
- For the first card, we can choose any of the 52 cards.
- For the second card, we need to choose a card from the 48 remaining cards that is not of the same rank as the first card.
So, the number of favorable outcomes = 52 * 48 = 2,496.
Therefore, the probability of not drawing a pair = favorable outcomes / total outcomes = 2,496 / 2,652 ≈ 0.9420 (or about 94.20%).
I hope this helps! Let me know if you have any further questions.
To find the probability of each event, we need to determine the number of favorable outcomes (the desired outcomes) and divide it by the number of possible outcomes.
In this case, there are 52 cards in a deck, and we're selecting two cards without replacement. So for the first card drawn, there are 52 possible choices, and for the second card drawn, there are 51 possibilities remaining.
Now let's find the favorable outcomes for each event:
a. Two cards of the same suit are drawn:
There are four suits in a deck (hearts, diamonds, clubs, and spades). To calculate the probability of drawing two cards of the same suit, we need to count the number of favorable outcomes.
For the first card drawn, there are 13 cards of the same suit (e.g., 13 hearts, 13 diamonds, etc.). For the second card drawn, there are 12 cards of the same suit remaining.
Therefore, the number of favorable outcomes is 13 * 12 = 156.
The total number of possible outcomes is 52 * 51 = 2652.
So the probability of drawing two cards of the same suit is 156/2652 ≈ 0.0588, or approximately 5.88%.
b. A pair is not drawn:
To calculate the probability of not drawing a pair, we need to count the number of favorable outcomes.
Since there are 52 cards in a deck, the first card can be any of the 52 cards. For the second card, there are 48 remaining cards that are different from the first card (the 3 other cards of the same value are the only ones that would form a pair).
Therefore, the number of favorable outcomes is 52 * 48 = 2496.
The total number of possible outcomes is still 52 * 51 = 2652.
So the probability of not drawing a pair is 2496/2652 ≈ 0.9408, or approximately 94.08%.
I hope this helps you understand how to calculate the probability in these situations! Let me know if you have any further questions.
consider the deck either as 4 suits of 13, or as 13 suits of 4.
P(same suit) = 1 * 12/51
P(~pair) = 1-P(pair) = 1 - 1 * 3/51