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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 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 32684
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 32292
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 28790
Q:

A bag contains 50 tickets numbered 1,2,3,4......50 of which five are drawn at random and arranged in ascending order of magnitude.Find the probability that third drawn ticket is equal to 30.

A) 551/15134 B) 1/2
C) 552/15379 D) 1/9
 
Answer & Explanation Answer: A) 551/15134

Explanation:

Total number of elementary events = 50C5
Given,third ticket =30

 

 

 

=> first and second should come from tickets numbered 1 to 29 = 29C2 ways and remaining two in 20C2 ways.

 

 

 

Therfore,favourable number of events = 29C2*20C2

 

 

 

Hence,required probability = 29C2*20C2/50C5 =551 / 15134

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72 28039
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 27997
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 27784
Q:

A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

A) 23/42 B) 19/42
C) 7/32 D) 16/39
 
Answer & Explanation Answer: B) 19/42

Explanation:

A red ball can be drawn in two mutually exclusive ways

 (i) Selecting bag I and then drawing a red ball from it.

 

(ii) Selecting bag II and then drawing a red ball from it.

 

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore 

P(E1) = 1/2  and  P(E2) = 1/2

Now, PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7

  PAE2  = Probability of drawing a red ball when the second bag has been selected = 2/6

 Using the law of total probability, we have 

 P(red ball) = P(A) = PE1×PAE1+PE2×PAE2 

 

                          = 12×47+12×26=1942

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64 27281
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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36 26943