Question

Question #59d64

Answer 1

The probability is 30.8%

Explanation:

There are 52 cards in all, and 13 are spades. In addition, 4 are aces, but one of these aces is also a spade! We won't count it twice.

So, in total, we have 13 + 3 = 16 cards that are either spade or ace (including one that is both spade and ace, which I assume is to be included in the sample of "successful choices").

Probability = (successful events) / (all events) = `16 -: 52 = 0.308` or 30.8 %

Note: If "either a spade or an ace" is not meant to include the card that is both spade and ace, then reduce the number of successful events to 15.

Answer 2

`4/13`

Explanation:

`P(AuuB)=P(A)+P(B)-P(AnnB)`

Let`" "A=`a spade`=>P(A)=13/52`

`B=`an Ace`=>P(B)=4/52`

`"also "P(AnnB)=1/52`
( ie the probability of the ace of spades)

`P(AuuB)=13/52+4/52-1/52`

`=16/52=4/13`