Probabilities

1. FINDING
PROBABILITY
OF AN EVENT
FINDING PROBABILITY OF AN EVENT
USING COMBINATION RULE
04
1

2. PROBLEM 9
Two balls are drawn together at random from a bag containing 4 red and 2 black balls. How
many outcomes are possible?
Red Black Total Selected
4 2 6 2
1 2 1 3 1 4 1 1 1 2
2 2 2 1 2 2
3 3 1 3 2
4 1 4 2
1
3 4
4
2
Total outcomes are = 15
We can find total outcomes by using Combination Rule
6C2 = 15
Amjad Ali
2

3. PROBLEM 10
Six white balls and four black balls, which are indistinguishable apart from color, are placed in a
bag. If six balls are taken from the bag, find the probability of their being three white balls and
three black
White Black Total
6 4 10
Six balls are selected at random n(S) = 10C6 = 210
A: Selecting 3 white and 3 black balls n(A) = 6C3 . 4C3 = 80
P(A) =
𝒏(𝐀)
𝒏(𝐒)
=
𝟖𝟎
𝟐𝟏𝟎
=
𝟖
𝟐𝟏
PROBLEM 11
Four items are taken at random from a box of 12 items and inspected. The box is rejected if
more than one item is found to be faulty. If there are three faulty items in the box, find the
probability that the box is accepted.
Faulty Good Total
3 9 12
Four items are selected at random n(S) = 12C4 = 495
A: Accepting the box of items [It is accepted if no faulty & 4 good OR 1 faulty & 3 good items]
n(A) = 3C0 . 9C4 + 3C1 . 9C3 = 378
P(A) =
𝒏(𝐀)
𝒏(𝐒)
=
𝟑𝟕𝟖
𝟒𝟗𝟓
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3

4. PROBLEM 12
An employer wishes to hire three people from a group of 15 applicants, 8 men and 7 women, all
of whom are equally qualified to fill the position. If he selects the three at random, what is the
probability that (i) all three will be men (ii) at least one will be a women?
Men Women Total
8 7 15
Three persons are selected at random n(S) = 15C3 = 455
(i) A: Selecting 3 men n(A) = 8C3 . 7C0 = 56
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟓𝟔
𝟒𝟓𝟓
=
𝟖
𝟔𝟓
(ii) B: Selecting at least one woman
(1woman and 2 men) OR (2 women and 1 man) OR (3 women and no man)
n(B) = 7C1 . 8C2 + 7C2 . 8C1 + 7C3 . 8C0 = 196 + 168 + 35 = 399
P(B) =
𝐧(𝐁)
𝐧(𝐒)
=
𝟑𝟗𝟗
𝟒𝟓𝟓
=
𝟓𝟕
𝟔𝟓
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4

5. 5
PROBLEM 13
From a pack of 52 playing cards, two are drawn at random. Find the probability that one is king
and the other is a queen.
King Queen Rest Total
4 4 44 52
Two cards are selected at random
n(S) = 52C2 = 1326
A: Selecting one king and one queen
n(A) = 4C1 . 4C1 .44C0 = 16
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟏𝟔
𝟏𝟑𝟐𝟔
=
𝟖
𝟔𝟔𝟑
PROBLEM 14
A set of 8 cards contains 1 joker. A and B are two players and A chooses 5 cards at random, B
taking the remaining 3 cards. What is the probability that A has joker?
Joker Rest Total
1 7 8
Player A chooses 5 cards at random
n(S) = 8C5 = 56
A: Player A has joker
n(A) = 1C1 . 7C4 = 35
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟑𝟓
𝟓𝟔
=
𝟓
𝟖
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6. PROBLEM 15
Of 12 eggs in a refrigerator, 2 are bad. From these 4 eggs are chosen at random to make a cake.
Find the following probabilities. (i) Exactly one is bad (ii) At least one is bad.
Bad Good Total
2 10 12
Four eggs are selected at random
n(S) = 12C4 = 495
(i) A: Selecting exactly one bad egg
n(A) = 2C1 . 10C3 = 240
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟐𝟒𝟎
𝟒𝟗𝟓
=
𝟏𝟔
𝟑𝟑
(ii) B: Selecting at least one bad egg
(1 bad & 3 good) OR (2 bad & 2 good)
n(B) = 2C1 . 10C3 + 2C2 . 10C2 = 285
P(B) =
𝐧(𝐁)
𝐧(𝐒)
=
𝟐𝟖𝟓
𝟒𝟗𝟓
=
𝟏𝟗
𝟑𝟑
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6

7. PROBLEM 16
A certain carton has 3 bad eggs and 9 good eggs. An omelet is made of three eggs randomly
chosen from the carton. What is the probability that there are (i) no bad eggs (ii) at least one
bad egg (iii) exactly one bad egg in the omelet?
Bad Good Total
3 9 12
Three eggs are selected at random
n(S) = 12C3 = 220
(i) A: Selecting no bad egg
n(A) = 3C0 . 9C3 = 84
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟖𝟒
𝟐𝟐𝟎
=
𝟐𝟏
𝟓𝟓
(ii) B: Selecting at least one bad egg
[(1 bad & 2 good) OR (2 bad & 1 good) OR (3 bad & no good)
n(B) = 3C1 . 9C2 + 3C2 . 9C1 + 3C0 . 9C3 = 136
P(B) =
𝐧(𝐁)
𝐧(𝐒)
=
𝟏𝟑𝟔
𝟐𝟐𝟎
=
𝟑𝟒
𝟓𝟓
(iii) C: Selecting exactly one bad egg
n(C) = 3C1 . 9C2 = 108
P(C) =
𝐧(𝐂)
𝐧(𝐒)
=
𝟏𝟎𝟖
𝟐𝟐𝟎
=
𝟐𝟕
𝟓𝟓
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7

8. PROBLEM 17
Three distinct integers are chosen at random from first 20 positive integers. Compute the
probability that (i) their sum is even (ii) their product is even
Even Odd Total
10 10 20
Three integers are selected at random
n(S) = 20C3 = 1140
(i) A: Sum of numbers is even
[(3 even & no odd) OR (1 even & 2 odd)]
n(A) = 10C3 . 10C0 + 10C1 . 10C2 = 570
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟓𝟕𝟎
𝟏𝟏𝟒𝟎
=
𝟏
𝟐
(ii) B: Product of numbers is even
Note: Product will be even if there is at least one even number among the three selected
[(1 even & 2 odd) OR (2 even & 1 odd) OR (3 even & no odd)]
n(B) = 10C1 . 10C2 + 10C2 . 10C1 + 10C3 . 10C0 = 1020
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟏𝟎𝟐𝟎
𝟏𝟏𝟒𝟎
=
𝟏𝟕
𝟏𝟗
Amjad Ali
8

9. PROBLEM 18
A box contains 4 red, 4 white, and 5 green balls. 3 balls are drawn from the bag together. Find
the probability that they may be (i) all of different colors (ii) all of same colors
Red White Green Total
4 4 5 13
Three balls are selected at random
n(S) = 13C3 = 286
(i) A: Selecting three balls of different colors
[1 red & 1 white & 1 green]
n(A) = 4C1 . 4C1 . 5C1 =80
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟖𝟎
𝟐𝟖𝟔
=
𝟒𝟎
𝟏𝟒𝟑
(ii) B: Selecting all balls of same color
[3 red OR 3 white OR 3 green]
n(B) = 4C3 . 9C0 + 4C3 . 9C0 + 5C3 . 8C0 = 18
P(B) =
𝐧(𝐁)
𝐧(𝐒)
=
𝟏𝟖
𝟐𝟖𝟔
=
𝟗
𝟏𝟒𝟑
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9

10. PROBLEM 19
A bag contains 14 identical balls, 4 of which are red, 5 black, and 5 white. Six balls are selected
from the bag. Find the probability that (i)3 are red (ii)at least two are white.
Red Black White Total
4 5 5 14
Six balls are selected at random
n(S) = 14C6 = 3003
(i) A: Selecting three red balls
[3 red & 3 non-red]
n(A) = 4C3 . 10C3 = 480
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟒𝟖𝟎
𝟑𝟎𝟎𝟑
=
𝟏𝟔𝟎
𝟏𝟎𝟎𝟏
(ii) B: Selecting at least two white balls
[(2 white & 4 non-white) OR (3 white & 3 non-white) OR (4 white & 2 non-white) OR
(5 white & 1 non-white)]
n(B) = 5C2 . 9C4 + 5C3 . 9C3 + 5C4 . 9C2 + 5C5 . 9C1 = 2289
P(B) =
𝐧(𝐁)
𝐧(𝐒)
=
𝟐𝟐𝟖𝟗
𝟑𝟎𝟎𝟑
=
𝟏𝟎𝟗
𝟏𝟒𝟑
Red Non-red Total
4 10 14
White Non-white Total
5 9 14
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10

11. PROBLEM 20
Three applicants are to be selected at random out of 4 boys and 6 girls. Find the probability of
selecting (i) all girls (ii) all boys (iii)at least one boy.
Boys Girls Total
4 6 10
Three applicants are selected at random
n(S) = 10C3 = 120
(i) A: Selecting 3 girls
n(A) = 6C3 . 4C0 = 20
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟐𝟎
𝟏𝟐𝟎
=
𝟏
𝟔
(ii) B: Selecting three boys
n(B) = 4C3 . 6C0 = 6
P(B) =
𝐧(𝐁)
𝐧(𝐒)
=
𝟒
𝟏𝟐𝟎
=
𝟏
𝟑𝟎
(iii) C: Selecting at least one boy [1boy & 2 girls OR 2 boys & 1 girl OR 3boys & no girl]
n(C) = 4C1 . 6C2 + 4C2 . 6C1 + 4C3 . 6C0 = 100
P(C) =
𝐧(𝐂)
𝐧(𝐒)
=
𝟏𝟎𝟎
𝟏𝟐𝟎
=
𝟓
𝟔
Amjad Ali
11

12. PROBLEM 21
A firm buys three shipments of parts each month. The purchasing agent selects at random from
among four in-state suppliers and six out-of-state suppliers. What is the probability that the
orders are placed with the (i) in-state suppliers only? (ii) the out-of-state suppliers only? (iii) at
least one in-state supplier?
In state Out of State Total
4 6 10
Three are selected at random
n(S) = 10C3 = 120
(i) A: Selecting 3 in-state suppliers
n(A) = 4C3 . 6C0 = 4
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟒
𝟏𝟐𝟎
=
𝟏
𝟑𝟎
(ii) B: Selecting 3 out-of-state suppliers
n(B) = 4C0 . 6C3 = 20
P(B) =
𝐧(𝐁)
𝐧(𝐒)
=
𝟐𝟎
𝟏𝟐𝟎
=
𝟏
𝟔
(iii) C: Selecting at least one in-state supplier
[1 instate & 2 out of state OR 2 instate & 1 out of state OR 3 in state & no out of state]
n(C) = 4C1 . 6C2 + 4C2 . 6C1 + 4C3 . 6C0 = 100
P(C) =
𝐧(𝐂)
𝐧(𝐒)
=
𝟏𝟎𝟎
𝟏𝟐𝟎
=
𝟓
𝟔 Amjad Ali
12

13. PROBLEM 22
From a group of 6 men and 8 women, 5 people are chosen at random. Find the probability that
there are more men chosen than woman.
Men Women Total
6 8 14
Five persons are selected at random n(S) = 14C5 = 2002
A: Selecting more men then women
(3 men & 2 women) OR (4 men & 1 woman) OR (5 men & no woman)
n(A) = 6C3 . 8C2 + 6C4 . 8C1 + 6C5 . 8C0 = 686
P(A) =
𝐧(𝐀)
𝐧(𝐒)
=
𝟔𝟖𝟔
𝟐𝟎𝟎𝟐
=
𝟒𝟗
𝟏𝟒𝟑
Amjad Ali
13

14. THANK
YOU
04
Amjad Ali
14