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, ES

P(E)=n(E)n(S)

6. Results on Probability :

i. P(S) = 1    ii. 0P(E)1   iii. P()=0

 

iv. For any events A and B we have : 

P(AB)=P(A)+P(B)-P(AB)

 

v. If A denotes (not-A), then P(A)=1-P(A)

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 29800
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 28319
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 27186
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 26583
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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37 25697
Q:

From a pack of 52 cards, 3 cards are drawn. What is the probability that one is ace, one is queen and one is jack?

A) 19/5525 B) 16/5525
C) 17/5525 D) 7/5525
 
Answer & Explanation Answer: B) 16/5525

Explanation:

Required probability:

 4C1*4C1*4C152C3=4*4*422100=165525

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27 25014
Q:

A word consists of 9 letters; 5 consonants and 4 vowels.Three letters are choosen at random. What is the probability that more than one vowel will be selected ?

A) 13/42 B) 17/42
C) 5/42 D) 3/14
 
Answer & Explanation Answer: B) 17/42

Explanation:

3 letters can be choosen out of 9 letters in 9C3 ways.

 

More than one vowels ( 2 vowels + 1 consonant  or  3 vowels ) can be choosen in (4C2*5C1)+4C3 ways

 

Hence,required probability = 4C2*5C1+4C39C3 = 17/42

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55 24488
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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44 24103