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 purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random and are placed in a box. If a coin is now picked from the box, find the probability of it being a rupee coin?

A) 4/7 B) 2/3
C) 1/2 D) 5/6
 
Answer & Explanation Answer: B) 2/3

Explanation:

Total coins 30

In that,

1 rupee coins 20

50 paise coins 10

Probability of total 1 rupee coins =  20C11

Probability that 11 coins are picked = 30C11

Required probability of a coin now picked from the box is 1 rupee = 20C11/30C11 = 2/3.

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17 11002
Q:

If an unbiased dice is rolled once, the odds in favour of getting a point which is multiple of 3 is:

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

Explanation:

Total number =6

Getting a 'multiple of 3' = 2 . So, probability = 2/6 =1/3

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15 10933
Q:

The probabilities that a student will receive an A, B, C or D grade are 0.4, 0.3 , 0.2 and 0.1 respectively. Find the probability that a student will receive Atleast B grade.

A) 0.21 B) 0.3
C) 0.7 D) None of these
 
Answer & Explanation Answer: C) 0.7

Explanation:

P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7

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43 10663
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 10661
Q:

If P(A)=4/9;then the odd against the event A is:

A) 4:9 B) 4:5
C) 5:4 D) 4:14
 
Answer & Explanation Answer: C) 5:4

Explanation:

Here,P(A)=4/9=p/(p+q)

P(odd against event A)=q/p=5/4

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11 10523
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=PABPB=2868=13

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15 10439
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 10375
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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