### Probability - Basic Concepts

• Understand basic terminology related to probability such as sample space, event, complement, boolean combination and dependency of multiple events.

### Solved Example:

#### 7-1-01

From a pack of regular playing cards, two cards are drawn at random. What is the probability that both cards will be Kings, if first card in NOT replaced?

Solution:
Given : Total number of cards = 52 and two cards are drawn at random. Number of kings in playing cards = 4 So the probability that both cards will be king is given by, $P = \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{1}{221}$

### Solved Example:

#### 7-1-02

Two dice are thrown. What is the probability that the sum of the numbers on the two dice is eight?

Solution:
We know that a dice has 6 faces and 6 numbers so the total number of cases (outcomes) = 6$\times$6 = 36 And total ways in which sum of the numbers on the dices is eight, $\left\{(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)\right\}$
So, the probability that the sum of the numbers eight is, p =$\dfrac{5}{36}$

### Solved Example:

#### 7-1-03

A box contains 5 black and 5 red balls. Two balls are randomly picked one after another form the box, without replacement. The probability for balls being red is:

Solution:
$\mathrm{Black\ balls} = 5, \mathrm{Red\ balls} = 5, \mathrm{Total\ balls} = 10$ Here, two balls are picked from the box randomly one after the other without replacement. So the probability of both the balls are red is: $P =\dfrac {5C_{0}\times 5C_{2}}{10C_{2}} =\dfrac {\dfrac {5!}{0!5!}\times \dfrac {5!}{3!2!}}{\dfrac {10!}{3!2!}} =\dfrac {1\times 10}{45} =\dfrac {2}{9}$
Alternate Method: The probability of drawing a red ball, $P_1 = \dfrac{5}{10} = \dfrac{1}{2}$
If ball is not replaced, then box contains 9 balls. So, probability of drawing the next red ball from the box. $P_2= \dfrac{4}{9}$ Hence, probability for both the balls being red is, $P = P_1 \times P_2 = \dfrac{1}{2} \times \dfrac{4}{9} = \dfrac{2}{9}$

### Solved Example:

#### 7-1-04

A single die is thrown twice. What is the probability that the sum is neither 8 nor 9?

Solution:
We know a die has 6 faces and 6 numbers so the total number of ways = 6$\times$6 = 36. And total ways in which sum is either 8 or 9 is 9, i.e. $(2,6), (3,6) (3,5) (4,4) (4,5) (5,4) (5,3) (6,2) (6,3)$
Total number of tosses when both the 8 or 9 numbers are not come = 36 - 9 = 27. Then probability of not coming sum 8 or 9 is, 36 = $\dfrac{27}{36}$ = $\dfrac{3}{4}$

### Solved Example:

#### 7-1-05

A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective ?

Solution:
Sample Space= Any two items can be selected from 100 items = $100_{C_2}$
Favourable case= Defective two items can be selected from 100 items = $20_{C_2}$ Probability of the required event is given by: $= \dfrac{20_{C_2}}{100_{C_2}} = \dfrac{\dfrac{20!}{18!2!}}{\dfrac{100!}{98!2!}} = \dfrac{20 \times 19}{100 \times 99} = \dfrac{19}{495}$

### Solved Example:

#### 7-1-06

A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is:

Solution:
The box contains:
Number of washers = 2, Number of nuts = 3, Number of bolts = 4
Total objects = 2 + 3 + 4 = 9
First two washers are drawn from the box which contain 9 items. So the probability of drawing 2 washers is, $P_{1} =\dfrac {2_{C_{2}}}{9_{C_{2}}} =\dfrac {1}{\left( \dfrac {9!}{7!2!}\right)} =\dfrac {7!2!}{9\times 8\times 7!} =\dfrac {2}{9\times 8} =\dfrac{1}{36}$ After this box contains only 7 objects and then 3 nuts drawn from it. So the probability of drawing 3 nuts from the remaining objects is, $P_{2} =\dfrac {3_{C_{3}}}{7_{C_{3}}} =\dfrac {1}{\left( \dfrac {7!}{4!3!}\right) } =\dfrac {4!3!}{7\times 6\times 5\times 4!} =\dfrac {1}{35}$ After this box contain only 4 objects, probability of drawing 4 bolts from the box, $P_{3}=\dfrac {4_{C_{4}}}{4_{C_{4}}} =\dfrac {1}{1} =1$
Therefore the required probability is, $P = P_1P_2P_3 = \dfrac{1}{36} \times \dfrac{1}{35} \times 1 = \dfrac{1}{1260}$

### Solved Example:

#### 7-1-07

A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is:

Solution:
No. of Red balls = 4, No. of Black ball = 6
3 balls are selected randomly one after another, without replacement.
1 red and 2 black balls are will be selected as following
$R B B: \dfrac{4}{10} \times \dfrac{6}{9} \times \dfrac{5}{8} = \dfrac{1}{6},$ $B R B: \dfrac{6}{10} \times \dfrac{4}{9} \times \dfrac{5}{8} = \dfrac{1}{6},$ $B B R: \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} = \dfrac{1}{6}$ Hence Total probability of selecting 1 red and 2 black ball is: $P = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{2}$

### Solved Example:

#### 7-1-08

In a company, 35% of the employees drink coffee, 40% of the employees drink tea and 10% of the employees drink both tea and coffee, what % of the employees drink neither tea nor coffee? (GATE Civil 2021)

Solution: Percent of employees drink neither tea nor coffee = 100 - 25 - 10 - 30 = 35