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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 fair coin is tossed 11 times. What is the probability that only the first two tosses will yield heads ?

A) (1/2)^11 B) (9)(1/2)
C) (11C2)(1/2)^9 D) (1/2)
 
Answer & Explanation Answer: A) (1/2)^11

Explanation:

Probability of occurrence of an event, 

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)


⇒ Probability of getting head in one coin = ½, 

⇒ Probability of not getting head in one coin = 1- ½ = ½, 

Hence, 

All the 11 tosses are independent of each other.

∴ Required probability of getting only 2 times heads =122×129=1211

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7 6477
Q:

What is the probability that a leap year selected at random, will contain 53 sundays?

A) 1/7 B) 1/3
C) 2/7 D) 4/7
 
Answer & Explanation Answer: C) 2/7

Explanation:

In a leap year,there are 366 days=52 weeks and 2 days

Remaining favourable 2 days can be sunday and monday or saturday and sunday

Exhaustive number of cases =7

Favourable number of cases =2

So,required probability=2/7

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0 6405
Q:

The probability that a number selected at random from the first 100 natural numbers is a composite number is  ?

A) 3/2 B) 2/3
C) 1/2 D) 34/7
 
Answer & Explanation Answer: A) 3/2

Explanation:

The number of exhaustive events = 100 C₁ = 100.

We have 25 primes from 1 to 100.

Number of favourable cases are 75.

Required probability = 75/50 = 3/2.

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13 6240
Q:

One card is drawn from a pack of 52 cards , each of the 52 cards being equally likely to be drawn. Find the probability that the card  drawn is neither a spade nor a king.

A) 0 B) 9/13
C) 1/2 D) 4/13
 
Answer & Explanation Answer: B) 9/13

Explanation:

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

 

Thus   n(E) = 13 + 3 = 16

 

Hence nE¯=52-16=36 

  

Therefore, PE=3652=913

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6 6237
Q:

Two unbiased coin are tossed. What is the probability of getting atmost one head?

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

Explanation:

Here, S={HH,HT,TH,TT}
Let E be the event of getting one head
E={TT,HT,TH}
P(E )= n(E)/n(S) =3/4

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5 6215
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 6117
Q:

The probabilities that drivers A, B and C will drive home safely after consuming liquor are 2 / 5, 3 / 7 and 3 / 4, respectively. What is the probability that they will drive home safely after consuming liquor ?

A) 3/70 B) 4/70
C) 9/70 D) 1/50
 
Answer & Explanation Answer: C) 9/70

Explanation:

Let A be the event of driver A drive safely after consuming liquor. 

 

Let B be the event of driver B drive safely after consuming liquor. 

 

Let C be the event of driver C drive safely after consuming liquor. 

 

P(A)=2/5,P(B)=3/7,P(C)=3/4

 

The events A, B and C are independent . Therefore,

 

PA∩B∩C=P(A)P(B)P(C)=25*37*34=970

 

Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9/70

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6 6096
Q:

A bag contains 2 red Roses, 4 yellow Roses and 6 pink Roses. Two roses are drawn at random. What is the probability that they are not of same color?

A) 1/6 B) 2/3
C) 14/33 D) 5/6
 
Answer & Explanation Answer: B) 2/3

Explanation:

Possible outcomes = (RY, YP, PR)

2C1 4C1 + 4C1 6C1 + 6C1 2C1

Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)

= 8 + 24 + 12/66

= 44/66

= 2/3.

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54 6093