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:

8 couples (husband and wife) attend a dance show "Nach Baliye' in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is :

A) 8/39 B) 15/39
C) 12/13 D) None of these
 
Answer & Explanation Answer: B) 15/39

Explanation:

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize) 

 

 = 1-16C1× 14C1×12C1×10C116C4=1539

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64 13333
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 28275
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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63 29535
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 24272
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 5883
Q:

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is :

A) 2/9 B) 1/9
C) 8/9 D) 7/9
 
Answer & Explanation Answer: B) 1/9

Explanation:

One person can select one house out of 3= 3C1 ways =3.

 

Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.

 

Therefore, probability that all thre apply for the same house is 1/9

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49 18141
Q:

In a bag which contains 40 balls, there are 18 red balls and some green and blue balls. If two balls are picked up from the bag without replacement, then the probability of the first ball being red and second being green is 3/26. Find the number of blue balls in the bag.

A) 16 B) 12
C) 10 D) 14
 
Answer & Explanation Answer: B) 12

Explanation:

Total balls = 40

Red balls = 18

Let green balls are x

Then, (18/40) × (

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49 4593
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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47 33410