Aptitude- Problems on Probability

Introduction:

Experiment:
An operation which can produce some well-defined outcome is called an experiment.

Random Experiment:

An experiment in which all possible out comes are known and the exact output cannot be predicted in advance is called a random experiment.
EX:
1) Rolling an unbiased dice.
2) Tossing a fair coin.
3) Drawing a card from a pack of well-shuffled cards .
4)Picking up a ball of certain colour from a bag containing
balls of different colours.

Details:

1) When we thrown a coin ,then either a Head(H) or a Tail(T) appears.

2)A dice is a solid cube ,having 6 faces,marked 1,2,3,4,5,6 respectively. When we throw a die ,the outcome is the number that appears on its upper face.

3)A pack of cards has 52 cards.It has 13 cards of each suit,namely spades,clubs,hearts and
diamonds. Cards of spades and clubs are balck cards.Cards of hearts and diamonds are red cards.
There are four honours of each suit.
These are Aces,Kings,queens and Jacks.
These are called Face cards.

Sample Space:
When we perform an experiment ,then the set of S of all
possible outcomes is called the Sample space .

EX:
1)In tossing a coin S= {H,T}.
2)If two coins are tossed then S= {HH,HT,TH,TT}.
3)In rolling a dice ,we have S={1,2,3,4,5,6}.

Event:
Any subset of a sample space is called an Event.Probability of occurrence of an Event:
Let S be the sample space.
Let E be the Event.
Then E cS i.e E is subset of S then
probability of E p(E) =n(E)/n(S).

Results on Probability:
1)P(S) =1.
2)0 < P(E) < 1 probability of an event lies between 0 and 1. Max value of probability of an event is one. 3)P(Ð¤)=0. 4)For any events A and B we have . P(AUB) =P(A) +P(B) -P(AnB). 5)If A denotes (not -A) then P(A) =1-P(A) P(A)+P(A) =1. Problems:

1)An biased die is tossed.Find the probability of getting a multiple of 3?

Sol: Here we have sample space S={1,2,3,4,5,6}.
Let E be the event of getting a multiple of 3.
Then E={3,6}.
P(E) =n(E)/n(S).
n(E) =2,
n(S) =6.
P(E) =2/6
P(E) =1/3.

2)In a simultaneous throw of a pair of dice,find the probability of getting a total more than 7?

Sol: Here we have sample space n(S) =6*6 =36.
Let E be the event of getting a total more than 7.
={(1,6),(2,5),(3,4),(4,3)(5,2),(6,1)(2,6),(3,5),(4,4),
(5,3),(6,2),(4,5),(5,4),
(5,5),(4,6),(6,4)}
n(E) =15
P(E) = n(E)/n(S)
= 15/36.
P(E) = 5/12.

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

Sol: Let S be the sample space.
Number of ways for drawing two balls out of 6 white and
4 red balls = 10C2
=10!/(8!*2!)
= 45.
n(S) =45.
Let E =event of getting both balls of the same colour.
Then
n(E) =number of ways of drawing ( 2balls out of 6) or
(2 balls out of 4).
= 6C2 +4C2
= 6!/(4!*2!) + 4!/(2! *2!)
= 6*5/2 +4 *3/2
=15+6 =21.
P(E) =n(E)/n(S) =21/45 =7/45.

4)Two dice are thrown together .What is the probability that the sum of the number on
the two faces is divisible by 4 or 6?

Sol: n(S) = 6*6 =36.
E be the event for getting the sum of the number on the two
faces is divisible by 4 or 6.
E={(1,3)(1,5)(2,4?)(2,2)(3,5)(3,3)(2,6)(3,1)(4,2)(4,4)
(5,1)(5,3)(6,2)(6,6)}
n(E) =14.
Hence P(E) =n(E)/n(S)
= 14/36.
P(E) = 7/18

5)Two cards are drawn at random from a pack of 52 cards What is the probability that either both are black or both are queens?
Sol: total number of ways for choosing 2 cards from
52 cards is =52C2 =52 !/(50!*2!)
= 1326.
Let A= event of getting bothe black cards.
Let B= event of getting bothe queens
AnB=Event of getting queens of black cards
n(A) =26C2.
We have 26 black cards from that we have to choose 2 cards.
n(A) =26C2=26!/(24!*2!)
= 26*25/2=325
from 52 cards we have 4 queens.
n(B) = 4C2
= 4!/(2!* 2!) =6
n(AnB) =2C2. =1
P(A) = n(A) /n(S) =325/1326
P(B) = n(B)/n(S) = 6/1326
P(A n B) = n(A n B)/n(S) = 1/1326
P(A u B) = P(A) +P(B) -P(AnB)
= 325/1326 + 6/1326 -1/1326
= 330/1326
P(AuB) = 55/221

6)Two diced are tossed the probability that the total score is a prime number?

Number of total ways n(S) =6 * 6 =36
E =event that the sum is a prime number.
Then E={(1,1)(1,2)(1,4)(1,6)(2,1)(2,3)(2,5)(3,2)(3,4)(4,1)
(4,3)(5,2)(5,6)(6,1)(6,5)}
n(E) =15
P(E) =n(E)/n(S)
= 15/36
P(E) = 5/12

7)Two dice are thrown simultaneously .what is the probability of getting two numbers whose product is even?

Sol : In a simultaneous throw of two dice ,we have n(S) = 6*6
= 36
E=Event of getting two numbers whose product is even
E={(1,2)(1,4)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,2)
(3,4)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,2)(5,4)(5,6)(6,1)
(6,2)(6,3)(6,4)(6,5)(6,6)}
n(E) = 27
P(E) = n(E)/n(S)
= 27 /36
P(E) =3/4
probability of getting two numbers whose product is even is
equals to 3/4.

8)In a lottery ,there are 10 prozes and 25 blanks.A lottery is drawn at random. what is the probability of getting a prize ?

Sol: By drawing lottery at random ,we have n(S) =10C1+25C1
= 10+25
= 35.
E =event of getting a prize.
n(E) =10C1 =10
out of 10 prozes we have to get into one prize .The number of
ways 10C1.
n(E) =10
n(S) =35
P(E) =n(E)/n(S)
=10/35
= 2/7
Probability is 2/7.

9)In a class ,30 % of the students offered English,20 % offered Hindi and 10 %offered Both.If a student is offered at random,what is the probability that he has offered English or Hindi?

Sol:English offered students =30 %.
Hindi offered students =20%
Both offered students =10 %
Then only english offered students E =30 -10
=20 %
only Hindi offered students S =20 -10 %
= 10 %
All the students =100% =E +S +E or S
100 =20 +10 + E or S +E and S
Hindi or English offered students =100 -20-10-10
=60 %
Probability that he has offered English or Hindi =60/100= 2/5

10) A box contains 20 electricbulbs ,out of which 4 are defective,two bulbs are chosen at random from this box.What is the probability that at least one of these is defective ?

Sol: out of 20 bulbs ,4 bulbs are defective.
16 bulbs are favourable bulbs.
E = event for getting no bulb is defective.
n(E) =16 C 2
out of 16 bulbs we have to choose 2 bulbs randomly .so the number
of ways =16 C 2
n(E) =16 C2
n(S) =20 C 2
P(E) =16 C2/20C2
= 12/19
probability of at least one is defective + probability of one
is non defective =1
P(E) + P(E) =1
12/19 +P(E) =1
P(Eâ€™) =7/19

11)A box contains 10 block and 10 white balls.What is the probability of drawing two balls of the same colour?

Sol: Total number of balls =10 +10
=20 balls
Let S be the sample space.
n(S) =number of ways drawing 2 balls out of 20
= 20 C2
= 20 !/(18! *2!)
= 190.
Let E =event of drawing 2 balls of the same colour.
n(E) =10C2+ 10C2
= 2(10 C2)
= 90
P(E) =n(E)/n(S)
P(E) =90/190
= 9/19

12) A bag contains 4 white balls ,5 red and 6 blue balls .Three balls are drawn at random from the bag.What is the probability that all of them are red ?

Sol: Let S be the sample space.
Then n(S) =number of ways drawing 3 balls out of 15.
=15 C3.
=455
Let E =event of getting all the 3 red balls.
n(E) = 5 C3 =5C2
= 10
P(E) =n(E) /n(S) =10/455 =2/91.

13)From a pack of 52 cards,one card is drawn at random.What is the probability that the card is a 10 or a spade?
Sol: Total no of cards are 52.
These are 13 spades including tne and there are 3 more tens.
n(E) =13+3
= 16
P(E) =n(E)/n(S).
=16/52
P(E) =4/13.

14) 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 probabililty of wife’s selection is 1/5.What is the probabililty
that only one of them is selected?

Sol: let A =event that the husband is selected.
B = event that the wife is selected.
E = Event for only one of them is selected.
P(A) =1/7
and
p(B) =1/5.
P(A’) =Probability of husband is not selected is =1-1/7=6/7
P(B’) =Probaility of wife is not selected =1-1/5=4/7
P(E) =P[(A and B’) or (B and A’)]
= P(A and B’) +P(B and A’)
= P(A)P(B’) + P(B)P(A’)
= 1/7*4/5 + 1/5 *6/7
P(E) =4/35 +6/35=10/35 =2/7

15)one card is drawn at random from a pack of 52 cards.What is the probability that the card drawn is a face card?

Sol: There are 52 cards,out of which there 16 face cards.
P(getting a face card) =16/52
= 4/13

16) The probability that a card drawn from a pack of 52 cards will be a diamond or a king?

Sol: In 52 cards 13 cards are diamond including one king there are
3 more kings. E event of getting a diamond or a king.
n(E) =13 +3
= 16
P(E) =n(E) /n(S) =16/52
=4/13

17) Two cards are drawn together from apack of 52 cards.What is the probability that one is a spade and one is a heart ?

Sol: S be the sample space the n (S) =52C2 =52*51/2
=1326
let E =event of getting 2 kings out of 4 kings
n(E) =4C2
= 6
P(E) =n(E)/n(S)
=6/1326
=1/221

18) Two cards are drawn together from a pack of 52 cards.What is the probability that one is a spade and one is a heart?

Sol: Let S be the sample space then
n(S) =52C2
=1326
E = Event of getting 1 spade and 1 heart.
n(E) =number of ways of choosing 1 spade out of 13 and 1 heart out
of 13.
= 13C1*13C1 =169
P(E)= n(E)/n(S)
=169/1326 =13/102.

19) Two cards are drawn from a pack of 52 cards .What is the probability that either both are Red or both are Kings?

Sol: S be the sample space.
n(S) =The number of ways for drawing 2 cards from 52 cards.
n(S) =52C2
=1326
E1 be the event of getting bothe red cards.
E2 be the event of getting both are kings.
E1nE2 =Event of getting 2 kings of red cards.
We have 26 red balls.From 26 balls we have to choose 2 balls.
n(E1) =26C2
= 26*25/2
=325
We have 4 kings .out of 4 kings,we have to choosed 2 balls.
n(E2) =4C2
=6
n(E1nE2) =2C2 =1
P(E1) = n(E1)/n(S)
=325/1326
P(E2) =n(E2)/n(S)
=6/1326
P(E1nE2) =n(E1nE2)/n(S) =1/1326
P(both red or both kings) = P(E1UE2)
= P(E1) +P(E2)-P(E1nE2)
=325/1326 +6/1326 -1/1326
=330/1326 =55/221

1. Niraj Raj says:

A bag contains 4 balls.Two balls are drawn at random and are found to be white.Find the prob. that all balls are white

• jana says:

1/6

• jana says:

wrong question,bec we do’t know how many white ball’s in bag…if white ball’s are two then answer is:1/6

2. Niraj Raj says:

How many dice must be thrown so that there is better than even chance of obtaining a six

3. Suresh Kumar says:

In a college 15% of the students are girls, among them 80% of the girls went for picnic. Find the percentage of the girls among the total students went for picnic.

• N.alam says:

ANS:- 12%
let no of students=100
no. of girls=15
no of girls went to picnic=(80/100)*15=12
so, out of 100,12 girls went for picnic i.e. 12%