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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:

There are 10 Letters and 10 correspondingly 10 different Address. If the letter are put into envelope randomly, then find the Probability that Exactly 9 letters will at the Correct Address ?

A) 1/10 B) 1/9
C) 1 D) 0
 
Answer & Explanation Answer: D) 0

Explanation:

As we know we have 10 letter and 10 different address and one more information given that exactly 9 letter will at the correct address....so the remaining one letter automatically reach to their correct address
P(E) = favorable outcomes /total outcomes
Here favorable outcomes are '0'.
So probability is '0'.

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5 8124
Q:

The odds favouring the event of a person hitting a target are 3 to 5. The odds against the event of another person hitting the target are 3 to 2. If each of them fire once at the target, find the probability that both of them hit the target.

A) 1/20 B) 4/20
C) 1/20 D) 3/20
 
Answer & Explanation Answer: D) 3/20

Explanation:

Let A be the event of first person hitting the target,

P(A) =33+5=38  (odd in favour)

Let B be the event of Second person hitting a target.

P(B)=23+2=25 (odd against)

Since both events are independent and both will hit the target so,

P(A∩B)= P(A)P(B)=38×25=320

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3 7751
Q:

The probability that a card drawn at random from the pack of playing cards may be either a queen or an ace is:

A) 9/13 B) 11/13
C) 2/13 D) None of these
 
Answer & Explanation Answer: C) 2/13

Explanation:

Number of queen cards = 4

Number of ace cards = 4

 P(either a queen or ace) = 4/52 + 4/52= 8/52 = 2/13

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3 7649
Q:

14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.

A) 11/379 B) 21/628
C) 24/625 D) 26/247
 
Answer & Explanation Answer: C) 24/625

Explanation:

Total no of ways = (14 – 1)! = 13!

Number of favorable ways = (12 – 1)! = 11!

 

So, required probability = 11!×3!13! = 39916800×66227020800 = 24625

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22 7632
Q:

You toss a coin AND roll a die. What is the probability of getting a tail and a 4 on the die?

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

Explanation:

Probability of getting a tail when a single coin is tossed =12
Probability of getting 4 when a die is thrown =16

Required probability  =(12)×(16)
= 1/12

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3 7496
Q:

The first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ?

A) 1/7 B) 8!
C) 7! D) 1/14
 
Answer & Explanation Answer: D) 1/14

Explanation:

The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14
Hence the answer is 1/14. 

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15 7482
Q:

The probabilities that A and B will tell the truth are 2 / 3 and 4 / 5 respectively . What is the probability that they agree with each other ?

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

Explanation:

Let A be the event of A will tell truth. B be the event of B tell truth

 

 P(A)=23,P(Ac)=1-P(A)=13

 

 

 

P(B)=45,P(Bc)=1-P(B)=15 

 

When both agree then they say true or they say false together, that is 

 

 A∩B or Ac∩Bc

 

 

 

Also these events will be mutually exclusive :

 

 P(A∩B)+P(Ac∩Bc)=P(A)P(B)P(C)=23*45+13*15=35

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6 7471
Q:

Two dice are rolled simultaneously. Find the probability of getting the sum of numbers on the on the two faces divisible by 3 or 4?

A) 3/7 B) 7/11
C) 5/9 D) 6/13
 
Answer & Explanation Answer: C) 5/9

Explanation:

Here n(S) = 6 x 6 = 36

E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)}

=> n(E)=20

Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.

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35 7284