Below are some solved simple exercises and
applications:
1) Find the probability of getting a
multiple of 2 at a die roll.
Solution: The number of outcomes that
are favorable to respective event is three (these are: {2}, {4}, {6}).
The number of equally possible outcomes is six, so the probability is
3/6 = 1/2 = 50%.
2) There are three pairs of socks of
different colors in a basket. Two socks are randomly extracted from the
basket. What is the probability of getting two socks of same color?
Solution:
The number of equally possible cases is the number of all 2-size
combinations of socks, namely, C(6, 2) = 15. The number of favorable
cases is three, because we have three pairs of socks having the same
color. Thus, the probability is 3/15 = 1/5 = 20%.
3) An urn contains four white balls and six
black balls. Two balls are drawn simultaneously. Find the probability of
the events: a) A – drawing two white balls; b) B – drawing two black
balls; c) C – drawing two balls of the same color.
Solution: The
number of possible cases is C(10, 2). a)
The number of cases that are favorable to event A is C(4, 2);
therefore,
.
b) Similarly,
.
c) We have
.
The events A and B are incompatible, so
.
4) Two dice, one red and one blue, are
rolled. Consider the events: A – occurrence of a number less than 4 on
the red die; B – occurrence of a number less than 3 on the blue die.
Find P(A or B).
Solution: The cases that are
favorable to A are {1}, {2} and {3}; therefore, P(A)
= 3/6. The cases that are favorable to B are {1} and {2};
therefore, P(B) = 2/6. The cases that are favorable to
A and B correspond to the ordered pairs (1, 1), (1, 2), (2, 1), (2,
2), (3, 1), (3, 2), totalling six, in a probability field where the
number of equally possible cases is 6
x 6 = 36. We
then have P(A and B) = 6/36. The requested
probability is
.
5) At a blackjack game, calculate the probability for a player
to get a total of twenty points from the first two cards (provided no
other cards are shown), if a 52-card deck is used.
Solution: The
variants totaling twenty points are of the type A + 9 or 10 + 10 (as a
value; that is, any 2-size combination of cards from 10, J, Q,
K). We have sixteen variants A + 9 (4 aces and 4 nines) and C(16,
2) = 120 variants 10 + 10 (all 2-size combinations of cards from the
sixteen cards with a value of 10). The number of all possible
distribution variants for two cards is C(52, 2)=1326. The probability is
then P = (16 + 120)/1326 = 68/663.
6) We have two urns, the first containing
three white balls and four black balls and the second three white balls
and five black balls. A ball is drawn from a randomly chosen urn. Find
the probability for the drawn ball to be white.
Solution: Denote
the events: A – the first urn is the chosen one; B – the
second urn is the chosen one; C – the drawn ball is white. A
and B form a complete system of events and P(A) =
P(B) = 1/2. We have P(C│A) =
3/7 and P(C│B) = 3/8. According to total
probability formula, we have: P(C) = P(A)P(C│A)
+ P(B)P(C│B) = (1/2)
x (3/7) + (1/2)
x (3/8) = 45/112
= 0.40178.
7) Five cards are drawn at once from a
32-card deck, containing cards from 7's up to aces. What is the
probability of the five cards containing at least one queen (Q)?
Solution:
Denoting by A the event to be measured the five extracted
cards contain at least one Q, we then calculate the probability of
the contrary event
–
the five extracted cards contain no Q.The equally possible
elementary events are the occurrences of 5-size combinations of cards from the 32, a total
of C(32, 5). The combinations that are favorable to event
have
the form (xyztv), with x, y, z, t,
v taking any card as value, except the four Q-cards. They
total C(32 – 4, 5) = C(28, 5). We then have:
.
For a broad range of applications in gambling, see
the
gambling section.
Author |
The author of this page is Catalin Barboianu
(PhD). Catalin is a games mathematician and problem gambling researcher,
science writer and consultant for the mathematical aspects of gambling
for the gaming industry and problem-gambling institutions.
Profiles:
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