Question

26. A standard deck of playing cards consists of 52 cards, half of which are black and half of which are red. If three cards are picked at random from the deck with replacement, what is the probability that either all three are black or all three are red?

Answer 1

Answer:

1/4

Explanation:

The probability of getting one black card out of a 52 deck of cards (where 26 are red and 26 are black) is: $\frac{26}{52}$, (remember that probability of an event = (total amount of favorable outcomes/ total amount of outcomes)  by simplifying this fraction we get: $\frac{26}{52} = \frac{13}{26}=\frac{1}{2}$.

Therefore, the probability of getting one black card out of a deck of 52  cards is 1/2.

Since the way of picking cards is done with replacement, for the second card, if we want to get another black one we would have a probability of 1/2.

Again, for the third card we would still have a probability of 1/2 of getting a black one.

We want to get a black one on the first time AND on the second AND on the third one, therefore we have to use the rule of multiplication, which tells us we have to multiply all three probabilities:

P(having 3 black cards) = $(\frac{1}{2}) (\frac{1}{2})(\frac{1}{2})= \frac{1}{8}$

Applying the same way of thinking that we used before but now for the red cards (which are also 26) we have that the probability of getting one red card is 1/2

Thus,

P(having 3 red cards) =  $(\frac{1}{2}) (\frac{1}{2})(\frac{1}{2})= \frac{1}{8}$

Since the problem is asking us to get either 3 black cards OR 3 red cards, we are going to use the rule of sum and we're going to sum up both probabilities.

P(having 3 black cards OR 3 red cards) = $\frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$

Thus, the probability that either all three are black or all three are red is 1/4