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:

A bag contains 3 black, 4 white and 5 red balls. One ball is drawn at random. Find the probability that it is either black or red ball:

A) 2/3 B) 1/4
C) 5/12 D) 1/2
 
Answer & Explanation Answer: A) 2/3

Explanation:

P(black ball)=3/12

P(red ball)=5/12

P(black or red)=3/12+5/12=2/3

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1 6676
Q:

In a throw of a coin, find the probability of getting a head.

A) 1 B) 1/2
C) 1/4 D) 0.1
 
Answer & Explanation Answer: B) 1/2

Explanation:

Here, s={H,T} and E={H}
P(E) = n(E)/n(S) = 1/2

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8 6636
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 6325
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 6307
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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4 6199
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 6164
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 6122
Q:

Find the range and mode of the data 17, 18, 28, 19, 16, 18, 17, 29, 18

A) 12 and 18 B) 13 and 18
C) 12 and 17 D) 11 and 17
 
Answer & Explanation Answer: B) 13 and 18

Explanation:
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