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:

From a pack of 52 cards, 3 cards are drawn. What is the probability that one is ace, one is queen and one is jack?

A) 19/5525 B) 16/5525
C) 17/5525 D) 7/5525
 
Answer & Explanation Answer: B) 16/5525

Explanation:

Required probability:

 4C1*4C1*4C152C3=4*4*422100=165525

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27 26387
Q:

A word consists of 9 letters; 5 consonants and 4 vowels.Three letters are choosen at random. What is the probability that more than one vowel will be selected ?

A) 13/42 B) 17/42
C) 5/42 D) 3/14
 
Answer & Explanation Answer: B) 17/42

Explanation:

3 letters can be choosen out of 9 letters in 9C3 ways.

 

More than one vowels ( 2 vowels + 1 consonant  or  3 vowels ) can be choosen in (4C2*5C1)+4C3 ways

 

Hence,required probability = 4C2*5C1+4C39C3 = 17/42

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55 26069
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

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44 25833
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

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67 22906
Q:

Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.

A) 19/34 B) 5/4
C) 20/34 D) 25/34
 
Answer & Explanation Answer: D) 25/34

Explanation:

The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl.

So the required probability:
=817*916+917*816+817*716

 

934+934+734
2534

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21 22252
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

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25 22055
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 

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97 21738
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

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43 20308