From a normal pack of 52 playing cards, one card is selected at random. How can I find the probability of the card being either a black card or an ace?

From a normal pack of 52 playing cards, one card is selected at random. How can I find the probability of the card being either a black card or an ace?

2 Answers
May 9, 2018

#p_"black or ace" =7/13#

Explanation:

If the card is selected at random, every single card has the same probability to be picked, so the probability of any specific card is:

#p = 1/52#

Consider the set of outcomes that satisfy this condition. This is the set of all black cards (including aces) plus the set of red aces.

Of the #52# possible outcomes then we have #26# that are black plus #2# red aces, and a total of #28# outcomes that satisfy the condition.

The probability of picking a card that is either black or an ace is then:

#p_"black or ace" = 28/52 =7/13#

May 10, 2018

# 7/13#.

Explanation:

Let, #A and B# denote, the events that the card selected is an Ace

and a Black-coloured, resp.

Then, the Reqd. Prob. is, #P(AuuB)#.

We know that, #P(AuuB)=P(A)+P(B)-P(AnnB)......(star)#.

For P(A), there are #52# cards in a normal pack, out which #1# can be

selected in #52# ways.

There are #4# aces in a pack, so, #1# can be chosen in #4# ways.

#:. P(A)=4/52..............................................................(star_1)#.

Similarly, #P(B)=26/52.....................................................(star_2)#.

Note that, #AnnB# denotes the event that the selected card is

an ace and black-coloured, i.e., a black-coloured ace.

Out of #52# cards, there are #2# black-coloured aces.

Clearly, #P(AnnB)=2/52..................................................(star_3)#.

From #(star),(star_1),(star_2) and (star_3)#, we have,

#"The Reqd. Prob."=4/52+26/52-2/52=28/52=7/13#, as already

derived by Respected Andrea S. !

Enjoy Maths.!