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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:

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
 
Answer & Explanation Answer: C) 3/4

Explanation:

Let A, B, C be the respective events of solving the problem and A , B, C be the respective events of not solving the problem. Then A, B, C are independent event

∴A, B, C are independent events

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

 PA=12, PB=23, PC= 34

∴ P( none  solves the problem) = P(not A) and (not B) and (not C)  

                  = PA∩B∩C 

                  = PAPBPC         ∵ A, B, C are Independent                       

                  =  12×23×34  

                  = 14  

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

                = 1-14= 3/4

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968 203020
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 3 or 5?

A) 1/2 B) 3/5
C) 9/20 D) 8/15
 
Answer & Explanation Answer: C) 9/20

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

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303 150241
Q:

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
 
Answer & Explanation Answer: B) 7/15

Explanation:

Let S be the sample space

 

Then n(S) = no of ways of drawing 2 balls out of (6+4) =10C2 10 =10*92*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)

 

                =6C2+4C2 = 6*52*1+4*32*1= 15+6 = 21

 

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

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752 129785
Q:

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
 
Answer & Explanation Answer: C) 55/221

Explanation:

We have n(s) =52C2 52 = 52*51/2*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) = 52*512*1 = 26C2 = 325, n(B)= 26*252*1= 4*3/2*1= 6  and  n(A∩B) = 4C2 = 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

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359 113165
Q:

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

A) 2/91 B) 1/22
C) 3/22 D) 2/77
 
Answer & Explanation Answer: A) 2/91

Explanation:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = 15C3  =15*14*133*2*1= 455.  

Let E = event of getting all the 3 red balls.

 n(E) = 5C3 = 5*42*1 = 10.

 => P(E) = n(E)/n(S) = 10/455 = 2/91.

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200 108583
Q:

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
 
Answer & Explanation Answer: A) 2/7

Explanation:

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

 

Total number of prizes, n(E) = 10

 P(E)=n(E)n(S)=1035=27

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263 102037
Q:

Two dice are tossed. The probability that the total score is a prime number is:

A) 5/12 B) 1/6
C) 1/2 D) 7/9
 
Answer & Explanation Answer: A) 5/12

Explanation:

Clearly, n(S) = (6 x 6) = 36.

Let 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 = 5/12.

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212 85869
Q:

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
 
Answer & Explanation Answer: A) 2/7

Explanation:

Probability_35.jpg

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