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:

Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is  ?

A) 16/19 B) 1
C) 3/2 D) 17/20
 
Answer & Explanation Answer: D) 17/20

Explanation:

n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
P(E U P) = 10/20 + 8/20 - 1/20 = 17/20

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Q:

A number X is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that |X|<2

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

Explanation:

X can take 7 values.
To get |X|+2) take X={−1,0,1}

=> P(|X|<2) = Favourable CasesTotal Cases = 3/7

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Q:

In a simultaneous throw of two dice, what is the probability of getting a doublet ?

A) 1/6 B) 1/3
C) 4/7 D) 4/5
 
Answer & Explanation Answer: A) 1/6

Explanation:

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

 

Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

 P(E)=n(E)n(S)=636=16

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Q:

If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.

A) 5/12 B) 7/12
C) 3/14 D) 1/12
 
Answer & Explanation Answer: D) 1/12

Explanation:

Total number of elementary events = 10C5

 

Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is 7C5.

 

So,required probability =7C510C5 = 1/12.

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Q:

A sequence a, ax, ax2, ......, axn, has odd number of terms. Then the median is

 

A) axn2+1 B) axn2-1
C) axn-1 D) axn2
 
Answer & Explanation Answer: D) axn2

Explanation:
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Q:

A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:

A) 123/897 B) 23/67
C) 7/44 D) 12/45
 
Answer & Explanation Answer: C) 7/44

Explanation:

Here, n(E) = 7C1×5C1×5C1

 

And,  n(S) = 12C1*11C1*10C1

P(S) = 7*6*512*11*10 = 7/44

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Q:

fifteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together ?

A) 3/91 B) 2/73
C) 1/91 D) 3/73
 
Answer & Explanation Answer: A) 3/91

Explanation:

In a circle of n different persons, the total number of arrangements possible = (n - 1)!
Total number of arrangements = n(S) = (15 – 1)! = 14 !
Taking three persons as a unit, total persons = 13 (in 4 units)
Therefore no. of ways for these 13 persons to around the circular table = (13 - 1)! = 12!
In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =
n(E) = 12! X 3!
Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14!
= 12!x3x2 / 14x13x12! = 6/14x13 = 3/91

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Q:

A bag contains 8 red and 4 green balls. Find the probability that two balls are red and one ball is green when three balls are drawn at random. 

A) 56/99 B) 112/495
C) 78/495 D) None of these
 
Answer & Explanation Answer: B) 112/495

Explanation:

nS= C4 12=495

 

nE= C28×C14=112

 

P(E)=112495

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