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 first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ?

A) 1/7 B) 8!
C) 7! D) 1/14
 
Answer & Explanation Answer: D) 1/14

Explanation:

The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14
Hence the answer is 1/14. 

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15 7938
Q:

The probability that a card drawn at random from the pack of playing cards may be either a queen or an ace is:

A) 9/13 B) 11/13
C) 2/13 D) None of these
 
Answer & Explanation Answer: C) 2/13

Explanation:

Number of queen cards = 4

Number of ace cards = 4

 P(either a queen or ace) = 4/52 + 4/52= 8/52 = 2/13

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3 7905
Q:

Two dice are rolled simultaneously. Find the probability of getting the sum of numbers on the on the two faces divisible by 3 or 4?

A) 3/7 B) 7/11
C) 5/9 D) 6/13
 
Answer & Explanation Answer: C) 5/9

Explanation:

Here n(S) = 6 x 6 = 36

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

=> n(E)=20

Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.

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35 7826
Q:

A person starting with 64 rupees and making 6 bets, wins three times and loses three times, the wins and loses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is:

A) A gain of Rs. 27 B) A loss of Rs. 37
C) A loss of Rs. 27 D) A gain of Rs. 37
 
Answer & Explanation Answer: B) A loss of Rs. 37

Explanation:

As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order).

The overall resultant will remain same.

So final amount with the person will be (in all cases):

64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27

Hence the final result is: 

64 − 27 37

A loss of Rs.37

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22 7821
Q:

You toss a coin AND roll a die. What is the probability of getting a tail and a 4 on the die?

A) 1/2 B) 1/12
C) 2/3 D) 3/4
 
Answer & Explanation Answer: B) 1/12

Explanation:

Probability of getting a tail when a single coin is tossed =12
Probability of getting 4 when a die is thrown =16

Required probability  =(12)×(16)
= 1/12

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3 7772
Q:

A number is selected from the numbers 1,2,3,4.......25.The probability for it to be divisible by 4 or 7 is:

A) 3/25 B) 9/25
C) 1/25 D) None of these
 
Answer & Explanation Answer: B) 9/25

Explanation:

Total numbers = 25

 Numbers divisible by 4 or 7 are 4, 7, 8, 12, 14, 16, 20, 21, 24 = 9

 

 The probability (divisible by 4 or 7) = 9/25

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3 7567
Q:

The variance of a set of data is 196. Then the standard deviation of the data is

 

A) +-14 B) 14
C) 96 D) 98
 
Answer & Explanation Answer: B) 14

Explanation:
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Q:

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize ?

A) 1/7 B) 2/7
C) 3/7 D) 2/5
 
Answer & Explanation Answer: B) 2/7

Explanation:

P(getting prize) = 10/ (10 + 25) =2/7

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