Question

A deck of cards with four suits; hearts, diamonds, spades, and clubs. you pick one card, put it back and thennpick another card. what is the probability that the first card is a diamond and the second card is not a diamond

Answer 1

1. First, let us find the probability that the first card is a diamond.

Now, since we are given that there are four suits and there are, assumably, an equal number of cards in each suit, we can say that the probability of choosing a diamond card is 1/4. We can also write this out as such, where D = Diamond:

Pr(D) = no. of diamond cards / total number of cards

There are 52 cards in a deck, and 13 cards of each suit, thus:

Pr(D) = 13/52 = 1/4

2. Now we need to calculate the probability of not choosing a diamond as the second card.

In many cases, when given a problem that requires you to find the probability of something not happening, it may be easier to set it out as such:

Pr(A') = 1 - Pr(A)

ie. Pr(A not happening, or not A) = 1 - Pr(A happening, or A)

This works because the total probability is always 1 (100%), and it makes sense that to find the probability of A not happening, we take the total probability and subtract the probability of A actually happening.

Thus, given that we have already calculated that the probability of choosing a Diamond is 1/4, we can now set this out as such:

Pr(D') = 1 - Pr(D)

Pr(D') = 1 - 1/4

Pr(D') = 3/4

3. Now we come to the final step. To find the probability of something and then something else happening, we must multiply the two probabilities together. Thus, given that Pr(D) = 1/4 and Pr(D') = 3/4, we get:

Pr(D)*Pr(D') = (1/4)*(3/4)

= 3/16

Thus, the probability of choosing a diamond as the first card and then not choosing a diamond as the second card is 3/16.