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:

A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

A) 10/21 B) 11/21
C) 1/2 D) 2/7
 
Answer & Explanation Answer: A) 10/21

Explanation:

Total number of balls = (2 + 3 + 2) = 7.

 

Let S be the sample space.

 

Then, n(S) = Number of ways of drawing 2 balls out of 7 =7C2 = 21

 

Let E = Event of drawing 2 balls, none of which is blue.

 

n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls =5C2 = 10

 

Therefore, P(E) = n(E)/n(S) = 10/ 21.

Report Error

View Answer Report Error Discuss

Filed Under: Probability

35 24368
Q:

A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a

king of heart is:

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

Explanation:

Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2.

P(E) =n(E)/n(S)=2/52=1/26.

Report Error

View Answer Report Error Discuss

Filed Under: Probability

7 23787
Q:

There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?

A) 1/2 B) 3/4
C) 4/7 D) 3/8
 
Answer & Explanation Answer: D) 3/8

Explanation:

Total cases of checking in the hotels = 4 x 4 x 4 = 64 ways.

Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways.

Required probability =24/64  = 3/8

Report Error

View Answer Report Error Discuss

Filed Under: Probability

44 23704
Q:

If two letters are taken at random from the word HOME, what is the probability that none of the letters would be vowels?

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

Explanation:

P(first letter is not vowel) = 24

 

P(second letter is not vowel) = 13

 

So, probability that none of letters would be vowels is = 24×13=16

Report Error

View Answer Report Error Discuss

Filed Under: Probability

67 22327
Q:

A letter is takenout at random from 'ASSISTANT'  and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

A) 35/96 B) 19/90
C) 19/96 D) None of these
 
Answer & Explanation Answer: B) 19/90

Explanation:

ASSISTANTAAINSSSTT

STATISTICSACIISSSTTT

Here N and C are not common and same letters can be A, I, S, T. Therefore

 Probability of choosing A =  2C19C1×1C110C1 = 1/45 

 Probability of choosing I = 19C1×2C110C1 = 1/45

Probability of choosing S = 3C19C1×3C110C1 = 1/10

Probability of choosing T = 2C19C1×3C110C1 = 1/15

Hence, Required probability =   145+145+110+115= 1990 

Report Error

View Answer Report Error Discuss

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

97 21181
Q:

An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.

A) 5/204 B) 1/204
C) 13/204 D) None of these
 
Answer & Explanation Answer: B) 1/204

Explanation:

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then 

 Required probability = PABC 

PA PBA PCAB

 Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When  a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

 PBA=317 

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

 PCAB =216=18

 Hence the required probability = 29×317×18=1204

Report Error

View Answer Report Error Discuss

Filed Under: Probability

25 20952
Q:

In a single throw of two dice , find the probability that neither a doublet nor a total of 8 will appear.

A) 7/15 B) 5/18
C) 13/18 D) 3/16
 
Answer & Explanation Answer: B) 5/18

Explanation:

n(S) = 36

 

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

 

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

 

nA=6, nB=5, nAB=1

 

Required probability = PAB

 

 = PA+PB-PAB

 

=  636+536-136 = 518

Report Error

View Answer Report Error Discuss

Filed Under: Probability

43 19918
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 19681