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, ES

P(E)=n(E)n(S)

6. Results on Probability :

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

 

iv. For any events A and B we have : 

P(AB)=P(A)+P(B)-P(AB)

 

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

Q:

What is the probability that a couple with four children have atleast one girl?

A) 0.0625 B) 0.9375
C) 0.5 D) 0.0257
 
Answer & Explanation Answer: B) 0.9375

Explanation:

Here,n = 4(children)

P(girl)= 0.5

P(of atleast one girl)= 1 - P(no girls)

                             = 1 - 0.0625 = 0.9375

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3 4247
Q:

Find the probability that it is either white or red ball when,one ball is drawn at random in a bag that contains 3 black,4 white and 5 red balls

A) 0.5 B) 0.75
C) 0.25 D) 0.8
 
Answer & Explanation Answer: B) 0.75

Explanation:

P(red ball) = 5/12

P(white ball) = 4/12

P(red or white ball) = 5/12 + 4/12 = 3/4 = 0.75

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1 4145
Q:

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

A) 21/46 B) 1/5
C) 3/25 D) 1/50
 
Answer & Explanation Answer: A) 21/46

Explanation:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

 

Then, n(S) = Number ways of selecting 3 students out of 25

 

                = 25C3  = 2300.

 

         n(E)= 10C1*15C2 = 1050. 

 

P(E) = n(E)/n(s) = 1050/2300 = 21/46

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1 4129
Q:

What is the probability that it is either a heart card or diamond card, when a card is drawn from a well shuffled standard pack of 52 playing cards?

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

Explanation:

P(heart cards)=13/52

P(diamond cards)=13/52

P(heart or diamond)=13/52+13/52=1/2

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2 3992
Q:

If the difference between the mode and median is 2, then the difference between the median and mean is:

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

Explanation:
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33 3905
Q:

Find the probability of selecting 2 woman when four persons are choosen at random from a group of 3 men, 2 woman and 4 children.

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

Explanation:

Out of 9 persons,4 can be choosen in 9C4 ways =126.

 

Favourable events for given condition = 2C2*7C2= 21.

 

So,required probability = 21/126 =1/6.

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1 3904
Q:

A five digit number is formed with the digits 0,1,2,3 and 4 without repetition. Find the chance that the number is divisible by 5.

A) 3/5 B) 1/5
C) 2/5 D) 4/5
 
Answer & Explanation Answer: B) 1/5

Explanation:
5 digit number = 5! = 120
Divisible by 5 then the last digit should be 0
Then the remaining position have the possibility = 4!
= 24
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25 3707
Q:

In a single throw with 2 dices, what is probability of neither getting an even number on one and nor a multiple of 3 on other?

A) 11/36 B) 25/36
C) 5/6 D) 1/6
 
Answer & Explanation Answer: B) 25/36

Explanation:

We first calculate the probability of getting an even number on one and a multiple of 3 on other,Here, n(s) = 6x6 = 36 and

E = (2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4)(3,6) (6,2)(6,4)

n(E) = 11P(E) = 11/36Required probability = 1-11/36 = 25/36

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17 3444