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 numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 15 ?

A) 6/19 B) 3/10
C) 7/10 D) 6/17
 
Answer & Explanation Answer: B) 3/10

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

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Q:

There are 26 balls marked with alphabetical order A to Z. What is the probability of selecting vowels listed balls? 

A) 1 B) 21/26
C) 5/26 D) 5
 
Answer & Explanation Answer: C) 5/26

Explanation:

We know that,

Total number of balls n(S) = 26

Number of vowels n(E) = 5

Hence, required probability = n(E)/n(S) = 5/26.

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44 4971
Q:

Three unbiased coins are tossed. What is the probability of getting at most two heads ?

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

Explanation:

Let S be the sample space.
Here n(S)= 23 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.

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Q:

If the standard deviation of a population is 10, what would be the population variance?

A) 100 B) 30
C) 5 D) 20
 
Answer & Explanation Answer: A) 100

Explanation:
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Q:

The mean of a distribution is 15 and the standard deviation is 5. What is the value of the coefficient variation?

A) 16.66% B) 66.66%
C) 33.33% D) 100%
 
Answer & Explanation Answer: C) 33.33%

Explanation:
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Q:

What is the probability of getting a sum 9 from two throws of a dice ?

A) 1/6 B) 1/2
C) 1/9 D) 3/4
 
Answer & Explanation Answer: C) 1/9

Explanation:

In two throws of a die, n(s)=(6 x 6)=36

 

let E= Event of geting a sum 9={(3,6),(4,5),(5,4),(6,3)}

 

P(E) = n(E)/n(S) = 4/36 = 1/9

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3 4838
Q:

When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?

A) 35/36 B) 17/36
C) 15/36 D) 1/36
 
Answer & Explanation Answer: A) 35/36

Explanation:

When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36

dice_thrown_simulataneously1532668754.png image

Required, the sum of the two numbers that turn up is less than 12

That can be done as n(E)

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

= 35

Hence, required probability = n(E)/n(S) = 35/36.

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Q:

Out of 3 girls and 6 boys a group of three members is to be formed in such a way that at least one member is a girl. In how many different ways can it be done?

A) 64 B) 84
C) 56 D) 20
 
Answer & Explanation Answer: A) 64

Explanation:

Total number of possible ways = 9C3 = 84 ways

Required atleast one girl in the group of three = total possible ways - ways in which none is girl

None of the members in the group is girl = 6C3 = 20

 

Therefore, number of ways that at least one member is a girl in the group of three

= 84 - 20

= 64 ways.

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