9.16 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICSProbability
Solution :
(a) Let A indicate the event of drawing 2 aces.
P A P(A) P A
A A(^) = ×
s
P(A) : drawing of an ace first
P A
A
: conditional probability of an ace at the second draw, given that the first was an ace.
Therefore,
P A 4 A 3P
A 52 A 51
P A 4 3 3 12 1
A 52 51 51 2652 221
(^) = (^) =
(^)
(^) = × = = =
(^)
(b) Let R indicate the event of drawing 2 red cards
P R P(R) P R
R R
26 25 650 25
52 51 2652 102
(^) = ×
(^)
= × = =
(c) Let E indicate the event of drawing an ace. Then the probability that at least an ace is drawn is
denoted by P(E). Probability of not drawing an ace :
P E P(E) P E
E E
48 47 2256 188
52 51 2652 221
(^) = ×
(^)
= × = =^
(^)
Therefor, probability of drawing at least one ace
1 188 33
= −221 221=
Example 21 :
The odds in favour of a certain event are 2 to 5 and the odds against another event independent of the
former are 5 to 6. Find the chance that one at least of the events will happen.
Solution :
The chance that the 1st event happens and the 2nd one does not happen
2 5 10
=7 11 77× =
The chance that the 1st event does not happen and the 2nd happens.
5 6 30
=7 11 77× =