Sawaal

 Probability Questions

Popular Latest Rated Important Formulae

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:

Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.

A) 19/34 B) 5/4
C) 20/34 D) 25/34
 
Answer & Explanation Answer: D) 25/34

Explanation:

The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl.

So the required probability:
=817*916+917*816+817*716

 

= 934+934+734
= 2534

Report Error

View Answer Report Error Discuss

Filed Under: Probability

21 19264
Q:

What is the probability of getting at least one six in a single throw of three unbiased dice?

A) 1/36 B) 91/256
C) 13/256 D) 43/256
 
Answer & Explanation Answer: B) 91/256

Explanation:

Find the number of cases in which none of the digits show a '6'.

i.e. all three dice show a number other than '6', 5×5×5=125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.

=216−125=91

The required probability = 91/256

Report Error

View Answer Report Error Discuss

Filed Under: Probability

27 18813
Q:

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is :

A) 2/9 B) 1/9
C) 8/9 D) 7/9
 
Answer & Explanation Answer: B) 1/9

Explanation:

One person can select one house out of 3= 3C1 ways =3.

 

Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.

 

Therefore, probability that all thre apply for the same house is 1/9

Report Error

View Answer Report Error Discuss

Filed Under: Probability

50 18239
Q:

An unbiased die is tossed.Find the probability of getting a multiple of 3.

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

Explanation:

Here S = {1,2,3,4,5,6}

Let E be the event of getting the multiple of 3

Then, E = {3,6}

P(E) = n(E)/n(S) = 2/6 = 1/3 

Report Error

View Answer Report Error Discuss

Filed Under: Probability

30 17447
Q:

If a card is drawn at random from a pack of 52 cards,what is the chance of getting a spade or ace?

A) 0.25 B) 5/13
C) 0.20 D) 4/13
 
Answer & Explanation Answer: D) 4/13

Explanation:

Number of spades in a standard deck of cards=13
Number of aces in a standard deck of cards=4
And,one of the aces is a spade.
So, 13 + 4 - 1 = 16 spades or aces to choose from.
Therfore,probabiltiy of getting a spade or an ace=16/52=4/13

Report Error

View Answer Report Error Discuss

Filed Under: Probability

18 16604
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

Report Error

View Answer Report Error Discuss

Filed Under: Probability

10 15540
Q:

The probability of success of three students X, Y and Z in the one examination are 1/5, 1/4 and 1/3 respectively. Find the probability of success of at least two.

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

Explanation:

P(X) = 15, P(Y) =14 , P(Z) = 13

 

Required probability:

 

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

 

=14*13*45+34*13*15+23*14*15+14*13*15

 

= 460+360+260+160= 1060= 16

Report Error

View Answer Report Error Discuss

Filed Under: Probability

17 15231
Q:

Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King

A) 1/52 B) 1/26
C) 1/13 D) 1/2
 
Answer & Explanation Answer: B) 1/26

Explanation:

Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)

 

=12C1*1C152C2+13C1*3C152C2

 

=12*252*51+13*3*252*51= 24+7852*51 = 126

Report Error

View Answer Report Error Discuss

Filed Under: Probability

26 15190