From a single deck of cards, you select a card, look at it, put it back in the deck and shuffle. If you do this 10 times, what is the probability of drawing exactly...
a) 1 Ace
b) 2 Aces
To calculate the probability of drawing a specific number of aces from a deck of cards, we need to know the total number of possible outcomes (the sample space) and the number of favorable outcomes (the desired outcome).
In this case, we know that the sample space consists of selecting one card from the deck ten times, which means we have 10 independent selections. The number of possible outcomes for each selection is 52 since there are 52 cards in a standard deck.
Now let's calculate the probability for each scenario:
a) Drawing exactly 1 Ace:
To calculate the probability of drawing one Ace, we need to consider that there are four Aces in the deck. In each of the ten selections, there is a probability of 4/52 of drawing an Ace, and a probability of 48/52 of drawing a non-Ace card. Since we want exactly 1 Ace, we need to consider one favorable outcome and nine non-favorable outcomes.
To calculate the probability, we use the binomial probability formula: P(X=k) = nCk * p^k * (1-p)^(n-k)
Here, n is the number of trials (10), k is the number of Aces drawn (1), p is the probability of drawing an Ace in one trial (4/52), and (1-p) is the probability of not drawing an Ace in one trial (48/52).
Using the formula, we have:
P(X=1) = 10C1 * (4/52)^1 * (48/52)^(10-1)
= 10 * (4/52) * (48/52)^9
You can simplify this expression further by calculating: 10 * (4/52) = 10/13.
Calculating the value of (48/52)^9, which is approximately 0.6991.
Finally, multiply these two results:
P(X=1) ≈ (10/13) * 0.6991 = 0.536
Therefore, the probability of drawing exactly 1 Ace is approximately 0.536 or 53.6%.
b) Drawing exactly 2 Aces:
Similarly, to calculate the probability of drawing two Aces, we need to consider that there are four Aces in the deck, and we want to select two of them in ten selections.
Using the same formula as before, we have:
P(X=2) = 10C2 * (4/52)^2 * (48/52)^(10-2)
Calculating this expression, we get:
P(X=2) ≈ (10! / (2! * 8!)) * (4/52)^2 * (48/52)^8
Simplifying further, we have:
P(X=2) ≈ (10 * 9 / (2 * 1)) * (4/52)^2 * (48/52)^8
P(X=2) ≈ (10 * 9 / 2) * (4/52)^2 * (48/52)^8
Now evaluate (4/52)^2 and (48/52)^8 separately:
(4/52)^2 ≈ 0.0025
(48/52)^8 ≈ 0.1861
When we substitute these values back into the expression, we get:
P(X=2) ≈ (10 * 9 / 2) * 0.0025 * 0.1861
P(X=2) ≈ 0.255
Therefore, the probability of drawing exactly 2 Aces is approximately 0.255 or 25.5%.