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:

Variance of the data 2, 4, 5, 6, 8, 17 is 23.33. Then the variance of 4, 8, 10, 12, 16, 34 will be

 

A) 11.66 B) 46.66
C) 93.3333 D) 4.83
 
Answer & Explanation Answer: C) 93.3333

Explanation:
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5 7384
Q:

Two dice are tossed. The probability that the total score is a prime number is :

A) 1/6 B) 1/2
C) 5/12 D) 3/4
 
Answer & Explanation Answer: C) 5/12

Explanation:

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4 7349
Q:

Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?

A) 1/4 B) 1/6
C) 1/8 D) 4
 
Answer & Explanation Answer: B) 1/6

Explanation:

Required probability is given by P(E) = n(E)n(S) = 14C2 = 16

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20 7288
Q:

A six-sided die with faces numbered 1 through 6 is rolled three times. What is the probability that the face with the number 6 on it will not face upward on all the three rolls?

A) 215/216 B) 1/216
C) 215/256 D) 1/52
 
Answer & Explanation Answer: A) 215/216

Explanation:

Probability of getting 6 at the top once =1/6

Probability of getting 6 at the top three times =1/6 x 1/6 x 1/6 =216

Probability of no getting 6 at he top any time = 1 - 1/216 = 215/216

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2 7156
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 7050
Q:

A number is selected at random from the numbers 1 to 30. What is the probability that it is divisible by either 3 or 7 ?

A) 1/30 B) 2/30
C) 1/25 D) 1/2
 
Answer & Explanation Answer: A) 1/30

Explanation:

Let A be event of selecting a number divisible by 3. B be the event of selecting a number divisible by 7.

  Proability tutorial

 A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 }, so n(A)=10
 B = { 7, 14, 21, 28 }, n(B)= 4 

Probability problems 

Since A and B are not mutually exclusive So :

Probability addition rule

Probability problems

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8 6900
Q:

Four persons are to be choosen from a group of 3 men, 2 women and 4 children. Find the probability of selecting 1 man,1 woman and 2 children.

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

Explanation:

Total number of persons = 9

 

Out of 9 persons 4 persons can be selected in 9C4 ways =126

 

1 man,1 woman and 2 children can be selected in 3C1*2C1*4C1 ways =36

So,required probability = 36/126 =2/7

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3 6844
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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