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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:

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

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48 36284
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.

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66 35489
Q:

Two brother X and Y appeared for an exam. The probability of selection of X is 1/7 and that of B is 2/9. Find the probability that both of them are selected.

A) 1/63 B) 1/14
C) 2/63 D) 1/9
 
Answer & Explanation Answer: C) 2/63

Explanation:

Let A be the event that X is selected and B is the event that Y is selected.

P(A) = 1/7,  P(B) = 2/9.

Let C be the event that both are selected.

P(C) = P(A) × P(B) as A and B are independent events: 

       = (1/7) × (2/9)  = 2/63

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30 34736
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

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148 34384
Q:

In a simultaneous throw of pair of dice. Find the probability of getting the total more than 7.

A) 1/2 B) 5/12
C) 7/15 D) 3/12
 
Answer & Explanation Answer: B) 5/12

Explanation:

Here n(S) = (6 x 6) = 36

Let E = event of getting a total more than 7
        = {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.

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47 30581
Q:

A speaks truth in 75% of cases and B in 80% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident

A) 30/100 B) 35/100
C) 45/100 D) 50/100
 
Answer & Explanation Answer: B) 35/100

Explanation:

Let   A = Event that A speaks the truth

B = Event that B speaks the truth 


Then P(A) = 75/100 = 3/4

P(B) = 80/100 = 4/5

P(A-lie) = 1-34= 1/4 

P(B-lie) = 1-45= 1/5

 

Now, A and B contradict each other =[A lies and B true] or [B true and B lies]

 = P(A).P(B-lie) + P(A-lie).P(B) 

 = 35*15+14*45=720  

 = 720*100= 35%

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64 29730
Q:

In a race, the odd favour of cars P,Q,R,S are 1:3, 1:4, 1:5 and 1:6 respectively. Find the probability that one of them wins the race.

A) 319/420 B) 27/111
C) 114/121 D) 231/420
 
Answer & Explanation Answer: A) 319/420

Explanation:

P(P)=14,P(Q)=15,P(R)=16,P(S)=17
All the events are mutually exclusive hence,

 

Required probability = P(P)+P(Q)+P(R)+P(S)

 

14+15+16+17=319420

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38 29668
Q:

A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

A) 10/21 B) 11/21
C) 1/2 D) 2/7
 
Answer & Explanation Answer: A) 10/21

Explanation:

Total number of balls = (2 + 3 + 2) = 7.

 

Let S be the sample space.

 

Then, n(S) = Number of ways of drawing 2 balls out of 7 =7C2 = 21

 

Let E = Event of drawing 2 balls, none of which is blue.

 

n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls =5C2 = 10

 

Therefore, P(E) = n(E)/n(S) = 10/ 21.

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37 29041