## Probability

Quantitative aptitude questions are asked in many competitive exams and placement exam. 'Probability' is a category in Quantitative Aptitude. Quantitative aptitude questions given here are extremely useful for all kind of competitive exams like Common Aptitude Test (CAT), MAT, GMAT, IBPS Exam, CSAT, CLAT, Bank Competitive Exams, ICET, UPSC Competitive Exams, CLAT, SSC Competitive Exams, SNAP Test, KPSC, XAT, GRE, Defense Competitive Exams, L.I.C/ G. I.C Competitive Exams, Railway Competitive Exam, TNPSC, University Grants Commission (UGC), Career Aptitude Test (IT Companies) and etc., Government Exams etc.

We have a large database of problems on "Probability" answered with explanation. These will help students who are preparing for all types of competitive examinations.

A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.

 A) 1/2 B) 7/15 C) 8/15 D) 1/9

Explanation:

Let S be the sample space

Then n(S) = no of ways of drawing 2 balls out of (6+4) = $\inline&space;{\color{Black}10C_{2}}$ =$\inline \fn_cm \frac{10\times 9}{2\times 1}$ =45

Let E = event of getting both balls of same colour

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

=$\inline&space;{\color{Black}6C_{2}+4C_{2}}$ = $\inline \fn_cm \frac{6\times 5}{2\times 1}+\frac{4\times 3}{2\times 1}$= 15+6 = 21

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

Subject: Probability - Quantitative Aptitude - Arithmetic Ability

67

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?

 A) 52/221 B) 55/190 C) 55/221 D) 19/221

Explanation:

We have n(s) = $\inline&space;{\color{Black}52C_{2}}$ = $\inline \fn_cm \frac{52\times 51}{2\times 1}$= 1326.

Let A = event of getting both black cards

B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = $\inline&space;{\color{Black}26C_{2}}$ = $\inline \fn_cm \frac{26\times 25}{2\times 1}$ = 325, n(B)= $\inline&space;{\color{Black}4C_{2}}$ = $\inline \fn_cm \frac{4\times 3}{2\times 1}$= 6  and  n(A∩B) = $\inline&space;{\color{Black}2C_{2}}$ = 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

Subject: Probability - Quantitative Aptitude - Arithmetic Ability

30

A man and his wife appear in an interview for two vacancies in the same post. The probability of husband's selection is (1/7) and the probability of wife's selection is (1/5). What is the probability that only one of them is selected ?

 A) 2/7 B) 1/7 C) 3/4 D) 4/5

Explanation:

Subject: Probability - Quantitative Aptitude - Arithmetic Ability

62

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

 A) 2/7 B) 5/7 C) 1/5 D) 1/2

Explanation:

Total number of outcomes possible, n(S) = 10 + 25 = 35

Total number of prizes, n(E) = 10

$\inline \fn_jvn p(E)= \frac{n(E)}{n(S)}=\frac{10}{35}=\frac{2}{7}$

Subject: Probability - Quantitative Aptitude - Arithmetic Ability

35

A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

 A) 1/4 B) 1/2 C) 3/4 D) 7/12

Explanation:

Let A, B, C be the respective events of solving the problem and $\inline \overline{A},\overline{B},\overline{C}$ be the respective events of not solving the problem. Then A, B, C are independent events

$\inline \therefore \bar{A},\bar{B},\bar{C}$ are independent events

Now,  P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

$\inline P(\bar{A})=\frac{1}{2},P(\bar{B})=\frac{2}{3} \: and\: P(\bar{C})=\frac{3}{4}$

$\inline \therefore$ P( none  solves the problem) = P(not A) and (not B) and (not C)

= $\inline P(\bar{A}\cap \bar{B}\cap \bar{C})$

= $\inline P(\bar{A})P(\bar{B})P(\bar{C})$             $\inline (\because \bar{A},\bar{B}\: and\: \bar{C}\: are\: independent)$                                  =  $\inline \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}$

= $\inline \frac{1}{4}$

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= $\inline 1-\frac{1}{4}=\frac{3}{4}$