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:

A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

A) 136/345 B) 17/87
C) 316/435 D) 158/435
 
Answer & Explanation Answer: C) 316/435

Explanation:

ns=C230

 

 Let A be the event of getting two oranges and 

 

 B be the event of getting two non-defective fruits.

 

 and A∩B be the event of getting two non-defective oranges

 

 ∴ PA=C220C230, PB=C222C230 and PA∩B=C215C230

 

 ∴PA∪B=PA+PB-PA∩B

 

= C220C230+C222C230-C215C230=316435

Report Error

View Answer Report Error Discuss

Filed Under: Probability

107 44261
Q:

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

A) 21/46 B) 1/5
C) 3/25 D) 1/50
 
Answer & Explanation Answer: A) 21/46

Explanation:

Let , S -  sample space        E - event of selecting 1 girl and 2 boys. 

Then, n(S) = Number ways of selecting 3 students out of 25 

                = 25C3 

                = 2300.

n(E) = 10C1×15C2 = 1050. 

P(E) = n(E)/n(s) = 1050/2300 = 21/46

Report Error

View Answer Report Error Discuss

Filed Under: Probability

106 42669
Q:

In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, what is the probability that he. has offered English or Hindi ?

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

Explanation:

3729a.jpg

Report Error

View Answer Report Error Discuss

Filed Under: Probability

73 40438
Q:

Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6.

A) 7/18 B) 14/35
C) 8/18 D) 7/35
 
Answer & Explanation Answer: A) 7/18

Explanation:

Clearly, n(S) = 6 x 6 = 36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.
Then,E = {(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}
n(E) = 14.
Hence, P(E) = n(E)/n(S) = 14/36 = 7/18

Report Error

View Answer Report Error Discuss

Filed Under: Probability

105 37149
Q:

I forgot the last digit of a 7-digit telephone number. If 1 randomly dial the final 3 digits after correctly dialing the first four, then what is the chance of dialing the correct number?

A) 1/999 B) 1/1001
C) 1/1000 D) 4/1000
 
Answer & Explanation Answer: C) 1/1000

Explanation:

It is given that last three digits are randomly dialled. then each of the digit can be selected out of 10 digits in 10 ways.
Hence required probability =1103 = 1/1000

Report Error

View Answer Report Error Discuss

Filed Under: Probability

28 35537
Q:

Four dice are thrown simultaneously. Find the probability that all of them show the same face.

A) 1/216 B) 1/36
C) 2/216 D) 4/216
 
Answer & Explanation Answer: A) 1/216

Explanation:

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:

 

= 6*6*6*6=64

 

n(S) = 64

 

Let X be the event that all dice show the same face. 

 

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

 

n(X) = 6

 

Hence required probability = n(X)n(S)=664 =1216

Report Error

View Answer Report Error Discuss

Filed Under: Probability

48 33588
Q:

What is the probability of getting 53 Mondays in a leap year?

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

Explanation:

1 year = 365 days . A leap year has 366 days

A year has 52 weeks. Hence there will be 52 Sundays for sure.

52 weeks = 52 x 7 = 364days

366 – 364 = 2 days

In a leap year there will be 52 Sundays and 2 days will be left.

These 2 days can be:

1. Sunday, Monday

2. Monday, Tuesday

3. Tuesday, Wednesday

4. Wednesday, Thursday

5. Thursday, Friday

6. Friday, Saturday

7. Saturday, Sunday

Of these total 7 outcomes, the favourable outcomes are 2.

Hence the probability of getting 53 days = 2/7

Report Error

View Answer Report Error Discuss

Filed Under: Probability

148 32338
Q:

A box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains exactly one defective bulb.

A) 5/12 B) 7/12
C) 3/14 D) 1/12
 
Answer & Explanation Answer: A) 5/12

Explanation:

Total number of elementary events = 10C5

 

Number of ways of selecting exactly one defective bulb out of 3 and 4 non-defective out of 7 is 3C1*7C4

 

So,required probability =3C1*7C4/10C5 = 5/12.

Report Error

View Answer Report Error Discuss

Filed Under: Probability

65 30307