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 Probability Questions

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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, E⊆S

∴P(E)=n(E)n(S)

6. Results on Probability :

i. P(S) = 1    ii. 0≤P(E)≤1   iii. P(∅)=0

 

iv. For any events A and B we have : 

P(A∪B)=P(A)+P(B)-P(A∩B)

 

v. If A denotes (not-A), then P(A)=1-P(A)

Q:

A coin is tossed 5 times. What is the probability that head appears an odd number of times?

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

Explanation:

The possible outcomes are as follows :

5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.

Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).

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27 16088
Q:

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

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

Explanation:

P(odd) = P (even) =12 1(because there are 50 odd and 50 even numbers)

 

Sum or the three numbers can be odd only under the following 4 scenarios:

 

Odd + Odd + Odd = 12*12*12= 18

 

Odd + Even + Even = 12*12*12=18

 

Even + Odd + Even = 12*12*12=18

 

Even + Even + Odd = 12*12*12 = 18

 

Other combinations of odd and even will give even numbers. 

 

Adding up the 4 scenarios above:

 

= 18+ 18+18+ 18 = 48 = 12

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10 15973
Q:

What is the probability that a leap year has 53 Saturdays and 52 Sundays ?

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

Explanation:

A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Friday, Saturday}

Required Probability = 1/7

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27 15653
Q:

A fair six-sided die is rolled twice. What is the probability of getting 4 on the first roll and not getting 6 on the second roll ?

A) 1/36 B) 5/36
C) 1/12 D) 1/9
 
Answer & Explanation Answer: B) 5/36

Explanation:

The two events mentioned are independent.

The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.

P(getting first 4) = 1/6

P(no second 6) = 5/6

Therefore P(getting first 4 and no second 6) = 1/6 x 5/6 = 5/36

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6 15307
Q:

In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples ?

A) 6/7 B) 19/21
C) 7/31 D) 5/21
 
Answer & Explanation Answer: D) 5/21

Explanation:

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ x ⁶C₁)

As remaining person can be any one among three couples left.

Required probability = (⁵C₂ x ⁶C₁)/¹⁰C₅
= (10 x 6)/252 = 5/21

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12 14974
Q:

A box contains 5 detective and 15 non-detective bulbs. Two bulbs are chosen at random. Find the probability that both the bulbs are non-defective.

A) 5/19 B) 3/20
C) 21/38 D) None of these
 
Answer & Explanation Answer: C) 21/38

Explanation:

n(S) = C220 = 190 

n(E) = C215 = 105 

Therefore, P(E) = 105/190 = 21/38

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11 14706
Q:

8 couples (husband and wife) attend a dance show "Nach Baliye' in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is :

A) 8/39 B) 15/39
C) 12/13 D) None of these
 
Answer & Explanation Answer: B) 15/39

Explanation:

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize) 

 

 = 1-16C1× 14C1×12C1×10C116C4=1539

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64 14490
Q:

A brother and a sister appear for an interview against two vacant posts in an office. The probability of the brother’s selection is 1/5 and that of the sister’s selection is 1/3. What is the probability that only one of them is selected?

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

Explanation:

Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) +  (Prob. that brother is not selected) × (Prob. that sister is selected)

 

= 15*23+45*13= 25

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14 14395