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:

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

A) 1/3 B) 3/5
C) 8/21 D) 7/21
 
Answer & Explanation Answer: A) 1/3

Explanation:

Total number of balls = (8 + 7 + 6) = 21.

Let E = event that the ball drawn is neither red nor green 

            = event that the ball drawn is blue.

n(E) = 7.

P(E) = n(E)/n(S) = 7/21 = 1/3.

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Q:

One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?

A) 3/13 B) 1/13
C) 3/52 D) 9/52
 
Answer & Explanation Answer: A) 3/13

Explanation:

Clearly, there are 52 cards, out of which there are 12 face cards.

P (getting a face card) = 12/52=3/13.

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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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Q:

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

A) 1/15 B) 1/221
C) 25/57 D) 35/256
 
Answer & Explanation Answer: B) 1/221

Explanation:

 Let S be the sample space

 

Then, n(S) = 52C2= 1326.

 

Let E = event of getting 2 kings out of 4.

 

n(E)  = 52*512*1= 6.

 4C2

=>4*32*1

P(E)=n(E)n(S)=61326=1221

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Q:

Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

A) 3/20 B) 29/34
C) 47/100 D) 13/102
 
Answer & Explanation Answer: D) 13/102

Explanation:

Let S be the sample space.

 

Then, n(S) = 52C2=(52 x 51)/(2 x 1) = 1326.

 

Let E = event of getting 1 spade and 1 heart.

 

n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = 13C1*13C1 = 169.

 

P(E) = n(E)/n(S) = 169/1326 = 13/102.

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Q:

Three unbiased coins are tossed. What is the probability of getting at most two heads?

A) 3/4 B) 7/8
C) 1/2 D) 1/4
 
Answer & Explanation Answer: B) 7/8

Explanation:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) =n(E)/n(S)=7/8.

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Q:

What is the probability of getting a sum 9 from two throws of a dice?

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

Explanation:

In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) =n(E)/n(S)=4/36=1/9.

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Q:

A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

A) 136/345 B) 17/87
C) 316/435 D) 158/435
 
Answer & Explanation Answer: C) 316/435

Explanation:

ns=C230

 

 Let A be the event of getting two oranges and 

 

 B be the event of getting two non-defective fruits.

 

 and AB be the event of getting two non-defective oranges

 

  PA=C220C230, PB=C222C230 and PAB=C215C230

 

 PAB=PA+PB-PAB

 

C220C230+C222C230-C215C230=316435

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