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, 

6. Results on Probability :

i. P(S) = 1

ii. 

iii. 

iv. For any events A and B we have : 

v. If  denotes (not-A), then 

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) = \inline {\color{Black}10C_{2}} = =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 {\color{Black}6C_{2}+4C_{2}} = = 15+6 = 21

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

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

 are independent events

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

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

                  = 

                  =                                                =  

                  = 

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

                = 

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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) = \inline {\color{Black}52C_{2}} = = 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 {\color{Black}26C_{2}} =  = 325, n(B)= \inline {\color{Black}4C_{2}} = = 6  and  n(A∩B) = \inline {\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

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83 26921
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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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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