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 Probability Questions

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FACTS  AND  FORMULAE  FOR  PROBABILITY  QUESTIONS

 

 

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

 

2. Random Experiment :An experiment in which all possible outcomes are know and the exact output cannot be predicted in advance, is called a random experiment.

Ex :

i. Tossing a fair coin.

ii. Rolling an unbiased dice.

iii. Drawing a card from a pack of well-shuffled cards.

 

3. Details of above experiments:

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

ii. 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.

iii. A pack of cards has 52 cards.

  • It has 13 cards of each suit, name Spades, Clubs, Hearts and Diamonds.
  • Cards of spades and clubs are black cards.
  • Cards of hearts and diamonds are red cards.

There are 4 honours of each unit. There are Kings, Queens and Jacks. These are all called face cards.

 

4. Sample Space: When we perform an experiment, then the set 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, the 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.

 

5. Probability of Occurrence of an Event : 

Let S be the sample and let E be an event.

Then, E⊆S

∴P(E)=n(E)n(S)

6. Results on Probability :

i. P(S) = 1    ii. 0≤P(E)≤1   iii. P(∅)=0

 

iv. For any events A and B we have : 

P(A∪B)=P(A)+P(B)-P(A∩B)

 

v. If A denotes (not-A), then P(A)=1-P(A)

Q:

In a bag which contains 40 balls, there are 18 red balls and some green and blue balls. If two balls are picked up from the bag without replacement, then the probability of the first ball being red and second being green is 3/26. Find the number of blue balls in the bag.

A) 16 B) 12
C) 10 D) 14
 
Answer & Explanation Answer: B) 12

Explanation:

Total balls = 40

Red balls = 18

Let green balls are x

Then, (18/40) × (

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Q:

A bag contains 5 red smileys, 6 yellow smileys and 3 green smileys. If two smileys are picked at random, what is the probability that both are red or both are green in colour?

A) 7 B) 1/7
C) 3/7 D) 0
 
Answer & Explanation Answer: B) 1/7

Explanation:

Given total number of smileys = 5 + 6 + 3 = 14

Now, required probability = 

5C214C2 + 3C214C2 = 1091 + 391 = 1391 = 17

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Q:

Out of sixty students, there are 14 who are taking Economics and 29 who are taking Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class ?

A) 8/15 B) 7/15
C) 1/15 D) 4/15
 
Answer & Explanation Answer: B) 7/15

Explanation:

Given total students in the class = 60
Students who are taking Economics = 24 and
Students who are taking Calculus = 32
Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4
Students who are taking calculus only = 32 - 4 = 28
probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.

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Q:

The probability that a bowler bowled a ball from a point will hit by the batsman is ¼. Three such balls are bowled simultaneously towards the batsman from that very point. What is the probability that the batsman will hit the ball ?

A) 37/64 B) 27/56
C) 11/13 D) 9/8
 
Answer & Explanation Answer: A) 37/64

Explanation:

Probability of not hitting the ball = 1- 1/4 =IBPS RRB Clerk Level Quiz : Quantitative Aptitude | 11 -09 - 17
Then, the probability that the batsman will hit the ball =IBPS RRB Clerk Level Quiz : Quantitative Aptitude | 11 -09 - 17

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Q:

One lady has 2 children, one of her child is boy, what is the probability of having both are boys ?

A) 1/3 B) 1/2
C) 2/3 D) 2/5
 
Answer & Explanation Answer: A) 1/3

Explanation:

In a family with 2 children there are four possibilities:
1) the first child is a boy and the second child is a boy (bb)
2) the first child is a boy and the second child is a girl (bg)
3) the first child is a girl and the second child is a boy (gb)
4) the first child is a girl and the second child is a girl (gg)

But already given that one child is boy. So we have three possibilities of (bb)(bg)(gb).
n(E)= both are boys=BB=1
n(S)= 3
Required probability P = n(E)/n(S) = 1/3.

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10 4761
Q:

In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box? 

A) 15 B) 18
C) 20 D) 24
 
Answer & Explanation Answer: C) 20

Explanation:

We know that, Total probability = 1

Given probability of black stones = 1/4

=> Probability of blue and white stones = 1 - 1/4 = 3/4

But, given blue + white stones =  9 + 6 = 15

Hence,

3/4 ----- 15

 1   -----  ?

=> 15 x 4/3 = 20.

 

Hence, total number of stones in the box = 20.

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16 4759
Q:

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 15 ?

A) 6/19 B) 3/10
C) 7/10 D) 6/17
 
Answer & Explanation Answer: B) 3/10

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

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6 4684
Q:

What is the probability of getting a sum 9 from two throws of a dice ?

A) 1/6 B) 1/2
C) 1/9 D) 3/4
 
Answer & Explanation Answer: C) 1/9

Explanation:

In two throws of a die, n(s)=(6 x 6)=36

 

let E= Event of geting a sum 9={(3,6),(4,5),(5,4),(6,3)}

 

P(E) = n(E)/n(S) = 4/36 = 1/9

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