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:

Tickets are numbered from 1 to 18 are mixed up together and then 9 tickets are drawn at random. Find the probability that the ticket has a number, which is a multiple of 2 or 3.

A) 1/3 B) 3/5
C) 2/3 D) 5/6
 
Answer & Explanation Answer: C) 2/3

Explanation:

S = { 1, 2, 3, 4, .....18 } 

=> n(S) = 18

 

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

 

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

 

 E3 =E1E2={6, 12, 18} 

=> n(E3) = 3

 

E=E1  E2 = E1+E2-E3 

=> n(E) = 9 + 6 - 3 =12

 where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

 

P(E)=n(E)n(S)=1218=23

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5 6242
Q:

What is the median of the following list of numbers

 

55, 53, 56, 59, 61, 69, and 31 ?

A) 55 B) 56
C) 59 D) 61
 
Answer & Explanation Answer: B) 56

Explanation:
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Q:

Consider 5 independent Bernoulli’s trials each with probability of success p. If the probability of at least one failure is greater than or equal to 31/32 , then p lies in the interval ?

A) [1, 32] B) (0, 1)
C) [1, 1/2] D) (1, 1/2]
 
Answer & Explanation Answer: C) [1, 1/2]

Explanation:

Probability of atleast one failure
= 1 - no failure > 31/32
= 1 - P5 > 31/32
=  P5<1/32
= p < 1/2
Also p > 0
Hence p lies in [0,1/2].

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5 5916
Q:

The probabiltiy that throw of two dice yields a total of 5 or 6 is:

A) 2/14 B) 5/18
C) 3/4 D) 1/4
 
Answer & Explanation Answer: B) 5/18

Explanation:

Total number of cases=36

Number of favourable cases(sum 5 or 6)=10

P(getting total of 5 or 6)=10/36=5/18

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5 5830
Q:

Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected?

A) 5/7 B) 1/5
C) 2/7 D) 2/35
 
Answer & Explanation Answer: C) 2/7

Explanation:

P( only one of them will be selected) = p[(E and not F) or (F and not E)] 

 = PEFFE 

 

PEPF+PFPE

 

 =17×45+15×67=27

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17 5821
Q:

Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails

A) 2 B) 7/8
C) 5/8 D) 1/2
 
Answer & Explanation Answer: D) 1/2

Explanation:

S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }

=> n(S) = 8

E = { HHH, HHT, HTH, THH }

=> n(E) = 4

P(E) = 4/8 = 1/2

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Q:

A room contains 3 brown, 5 black and 4 white chairs. Two chairs are picked and are put in the lawn. What is the probability that none of the chairs picked is white ?

A) 14/33 B) 14/55
C) 12/55 D) 13/33
 
Answer & Explanation Answer: A) 14/33

Explanation:

Total number of chairs = (3 + 5 + 4) = 12.

Let S be the sample space.

Then, n(s)= Number of ways of picking 2 chairs out of 12

12×11/2×66

Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.

=> 8C8×7/2×28

Therefore required probability = 28/66 = 14/33.

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15 5621
Q:

Two dice are rolled simultaneously. Find the probability of getting a total of 9.

A) 1/3 B) 1/9
C) 8/9 D) 9/10
 
Answer & Explanation Answer: B) 1/9

Explanation:

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

=> n(S) = 6 x 6 = 36

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

=> n(E) = 4

Therefore, P(E) = 4/36 = 1/9

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