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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:

If p:q are the odds in favour of an event,then the probability of that event is:

A) p/q B) p/(p+q)
C) q/(p+q) D) none
 
Answer & Explanation Answer: B) p/(p+q)

Explanation:

Total number of cases=p+q

Favourable cases=p

Probability of that event is=p/(p+q)

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7 10899
Q:

Four dice are thrown simultaneously. Find the probability that two of them show the same face and remaining two show the different faces.

A) 4/9 B) 5/9
C) 11/18 D) 7/9
 
Answer & Explanation Answer: B) 5/9

Explanation:

Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in C16, ways.

 

Now select two distinct number out of remaining 5 numbers which can be done in C25 ways. Thus these 4 numbers can be arranged in 4!/2! ways.

 

So, the number of ways in which two dice show the same face and the remaining two show different faces is 

 C16×C25×4!2!=720

 =>  n(E) = 720

 ∴PE=72064=59

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8 10789
Q:

A coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed. If 8 possible outcomes are equally likely. Find the probability that the die shows a number greater than 4, if it is known that the first throw of the coin results in a tail

A) 1/3 B) 2/3
C) 2/5 D) 4/15
 
Answer & Explanation Answer: A) 1/3

Explanation:

Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 }

 

Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,

 

 A = { T5, T6 }

 

 B = { T1, T2, T3, T4, T5, T6 }

 

Therefore, Required probability = PAB=PA∩BPB=2868=13

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15 10743
Q:

Three dice are thrown together.Find the probability of getting a total of atleast 6.

A) 103/216 B) 103/108
C) 103/36 D) 36/103
 
Answer & Explanation Answer: B) 103/108

Explanation:

Total number of events=6 x 6 x 6=216
Let A be the event of getting a total of atleast 6 and B denoted event of getting a total of less than 6 i.e.,3,4,5.

So,B={(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)}

Favourable number of cases=10

Therefore,P(B)=10/216

=> P(A) = 1 - P(B)

             = 1 - (10/216) = 103/108

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17 10498
Q:

An unbiased cubic die is thrown.What is the probabiltiy of getting a multiple of 3 or 4?

A) 1/12 B) 1/9
C) 3/4 D) 1/2
 
Answer & Explanation Answer: D) 1/2

Explanation:

Total numbers in a die=6

P(mutliple of 3) = 2/6 = 1/3

P(multiple of 4) = 1/6

P(multiple of 3 or 4) = 1/3 + 1/6 = 1/2

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4 10394
Q:

A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

A) 3/7 B) 4/7
C) 1/8 D) 3/4
 
Answer & Explanation Answer: B) 4/7

Explanation:

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) = 8 /14=4/7.

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6 10101
Q:

The letters of the word CASTIGATION is arranged in different ways randomly. What is the chance that vowels occupy the even places ?

A) 0.04 B) 0.043
C) 0.047 D) 0.05
 
Answer & Explanation Answer: B) 0.043

Explanation:

 Vowels are A I A I O, 

 

 C    A   S    T   I    G   A    T   I    O   N

 

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

 

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

 

n(E)=5P52!*2!(A,I are 2 times)*6P62!(T is 2 times)=21600

 

n(S)=11!2!*2!*2!(A, I are 2 times)=4989600

 

216004989600=0.043

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Exam Prep: GRE

7 10055
Q:

If x is chosen at random from the set {1,2,3,4} and y is to be chosen at random from the set {5,6,7}, what is the probability that xy will be even?

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

Explanation:
 

S ={(1,5),(1,6),(1,7),(2,5),(2,6),(2,7),(3,5),(3,6),(3,7),(4,5),(4,6),(4,7)}
Total element n(S)=12

xy will be even when even x or y or both will be even.
Events of x, y being even is E.
E ={(1,6),(2,5),(2,6),(2,7),(3,6),(4,5),(4,6),(4,7)}
n(E) = 8

P(E)=n(E)n(S)=812 = 4/3
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15 9777