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:

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

A) 3/4 B) 3/8
C) 5/16 D) 2/7
 
Answer & Explanation Answer: A) 3/4

Explanation:

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3,4),(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(E) = 27.

P(E) = n(E)/n(S) = 27/36 = 3/4.

Report Error

View Answer Report Error Discuss

Filed Under: Probability

44 20050
Q:

What is the probability of getting at least one six in a single throw of three unbiased dice?

A) 1/36 B) 91/256
C) 13/256 D) 43/256
 
Answer & Explanation Answer: B) 91/256

Explanation:

Find the number of cases in which none of the digits show a '6'.

i.e. all three dice show a number other than '6', 5×5×5=125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.

=216−125=91

The required probability = 91/256

Report Error

View Answer Report Error Discuss

Filed Under: Probability

27 19173
Q:

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is :

A) 2/9 B) 1/9
C) 8/9 D) 7/9
 
Answer & Explanation Answer: B) 1/9

Explanation:

One person can select one house out of 3= 3C1 ways =3.

 

Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.

 

Therefore, probability that all thre apply for the same house is 1/9

Report Error

View Answer Report Error Discuss

Filed Under: Probability

50 18539
Q:

The probability of success of three students X, Y and Z in the one examination are 1/5, 1/4 and 1/3 respectively. Find the probability of success of at least two.

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

Explanation:

P(X) = 15, P(Y) =14 , P(Z) = 13

 

Required probability:

 

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

 

=14*13*45+34*13*15+23*14*15+14*13*15

 

460+360+260+160106016

Report Error

View Answer Report Error Discuss

Filed Under: Probability

19 17979
Q:

An unbiased die is tossed.Find the probability of getting a multiple of 3.

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

Explanation:

Here S = {1,2,3,4,5,6}

Let E be the event of getting the multiple of 3

Then, E = {3,6}

P(E) = n(E)/n(S) = 2/6 = 1/3 

Report Error

View Answer Report Error Discuss

Filed Under: Probability

30 17845
Q:

If a card is drawn at random from a pack of 52 cards,what is the chance of getting a spade or ace?

A) 0.25 B) 5/13
C) 0.20 D) 4/13
 
Answer & Explanation Answer: D) 4/13

Explanation:

Number of spades in a standard deck of cards=13
Number of aces in a standard deck of cards=4
And,one of the aces is a spade.
So, 13 + 4 - 1 = 16 spades or aces to choose from.
Therfore,probabiltiy of getting a spade or an ace=16/52=4/13

Report Error

View Answer Report Error Discuss

Filed Under: Probability

18 17334
Q:

If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is  ?

A) 1 B) 1/2
C) 0 D) 3/5
 
Answer & Explanation Answer: B) 1/2

Explanation:

The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.

Report Error

View Answer Report Error Discuss

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

15 17055
Q:

Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King

A) 1/52 B) 1/26
C) 1/13 D) 1/2
 
Answer & Explanation Answer: B) 1/26

Explanation:

Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)

 

=12C1*1C152C2+13C1*3C152C2

 

=12*252*51+13*3*252*5124+7852*51 = 126

Report Error

View Answer Report Error Discuss

Filed Under: Probability

26 16488