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:

The letters of the word CASTIGATION is arranged in different ways randomly. What is the chance that vowels occupy the even places ?

A) 0.04 B) 0.043
C) 0.047 D) 0.05
 
Answer & Explanation Answer: B) 0.043

Explanation:

 Vowels are A I A I O, 

 

 C    A   S    T   I    G   A    T   I    O   N

 

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

 

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

 

n(E)=5P52!*2!(A,I are 2 times)*6P62!(T is 2 times)=21600

 

n(S)=11!2!*2!*2!(A, I are 2 times)=4989600

 

216004989600=0.043

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7 9235
Q:

If x is chosen at random from the set {1,2,3,4} and y is to be chosen at random from the set {5,6,7}, what is the probability that xy will be even?

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

Explanation:
 

S ={(1,5),(1,6),(1,7),(2,5),(2,6),(2,7),(3,5),(3,6),(3,7),(4,5),(4,6),(4,7)}
Total element n(S)=12

xy will be even when even x or y or both will be even.
Events of x, y being even is E.
E ={(1,6),(2,5),(2,6),(2,7),(3,6),(4,5),(4,6),(4,7)}
n(E) = 8

P(E)=n(E)n(S)=812 = 4/3

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15 9222
Q:

The probability that z lies between 0 and 3.01

A) 0.4987 B) 0.5
C) 0.9987 D) 0.1217
 
Answer & Explanation Answer: A) 0.4987

Explanation:

Probability between z = 0 and z = 3.01 is given by

P(0<z<3.01) = P(z<3.01) - P(z<0)

Reading from the z-table, we have

P(z<0) = 0.5

P(z<3.01) = 0.9987

Hence, P(0<z<3.01) = 0.9987 - 0.5 = 0.4987.

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Job Role: Analyst , Bank Clerk , Bank PO

37 8979
Q:

The probability of occurance of two two events A and B are 1/4 and 1/2 respectively. The probability of their simultaneous occurrance is 7/50. Find the probability that neither A nor B occurs.

A) 25/99 B) 39/100
C) 61/100 D) 17/100
 
Answer & Explanation Answer: B) 39/100

Explanation:

P ( neither A nor B) = PA and B  

 

=  PAB=  PAB1-PAB  

 

1-61100=39100

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8 8937
Q:

In a simultaneous throw of two dice, what is the probability of getting a total of 7 ?

A) 1/6 B) 1/4
C) 1/20 D) 3/4
 
Answer & Explanation Answer: A) 1/6

Explanation:

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16 8897
Q:

The probability that a card drawn from a pack of 52 cards will be a diamond or a king, is

A) 5/12 B) 4/13
C) 5/13 D) 3/14
 
Answer & Explanation Answer: B) 4/13

Explanation:

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8 8699
Q:

If P(A)=4/9;then the odd against the event A is:

A) 4:9 B) 4:5
C) 5:4 D) 4:14
 
Answer & Explanation Answer: C) 5:4

Explanation:

Here,P(A)=4/9=p/(p+q)

P(odd against event A)=q/p=5/4

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10 8694
Q:

In a charity show tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?

A) 0.25 B) 0.5
C) 0.75 D) 0.40
 
Answer & Explanation Answer: D) 0.40

Explanation:

250 numbers between 101 and 350 i.e. n(S)=250

n(E)=100th digits of 2 = 299−199 = 100

P(E)= n(E)/n(S) = 100/250 = 0.40

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3 8580