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 median of the following list of numbers

 

55, 53, 56, 59, 61, 69, and 31 ?

A) 55 B) 56
C) 59 D) 61
 
Answer & Explanation Answer: B) 56

Explanation:
Report Error

View Answer Report Error Discuss

Filed Under: Probability
Exam Prep: Bank Exams

1 3245
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.

Report Error

View Answer Report Error Discuss

Filed Under: Probability

1 3244
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
Report Error

View Answer Report Error Discuss

Filed Under: Probability

25 3164
Q:

From well shuffled standard pack of 52 playing cards,one card is drawn. What is the probability that it is either a red card or black card?

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

Explanation:

P(red cards)=26/52

P(black cards)=26/52

P(red or black cards)=26/52+26/52=1

Report Error

View Answer Report Error Discuss

Filed Under: Probability

1 2963
Q:

One card is drawn from deck of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the ball drawn is red.

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

Explanation:

Total number of elementary events = 52

 

There are 26 red cards,out of which one red card can be drawn in 26C1 ways =26.

 

So,required probability = 26/52 = 1/2

Report Error

View Answer Report Error Discuss

Filed Under: Probability

3 2838
Q:

Find the probability that one ball is white when two balls are drawn at random from a basket that contains 9 red, 7 white and 4 black balls.

A) 18/95 B) 18/190
C) 1/2 D) 91/190
 
Answer & Explanation Answer: D) 91/190

Explanation:

Total number of elementary events = 20C2 ways =190

 

There are 7 white balls out of which one white can be drawn in 7C1 ways and one ball from remaining 13 balls can be drawn in 13C1 ways.

 

So,favourable events = 7C1*13C1

 

Therfore,required probability = (7C1*13C1)/20C2=91/90

Report Error

View Answer Report Error Discuss

Filed Under: Probability

2 2747
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

Report Error

View Answer Report Error Discuss

Filed Under: Probability
Exam Prep: Bank Exams
Job Role: Bank PO , Bank Clerk

16 2630
Q:

The standard deviation of the set {10, 11, 12, 9, 8} is

A) 1 B) √2
C) 2 D) 2√2
 
Answer & Explanation Answer: B) √2

Explanation:
Report Error

View Answer Report Error Discuss

Filed Under: Probability
Exam Prep: Bank Exams

5 877