# Probability Questions

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, $E\subseteq S$

$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

6. Results on Probability :

i. P(S) = 1    ii. $0\le P\left(E\right)\le 1$   iii. $P\left(\varnothing \right)=0$

iv. For any events A and B we have :

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

v. If $\overline{)A}$ denotes (not-A), then $P\left(\overline{)A}\right)=1-P\left(A\right)$

Q:

A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

 A) 10/21 B) 11/21 C) 1/2 D) 2/7

Explanation:

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7 =$7C2$ = 21

Let E = Event of drawing 2 balls, none of which is blue.

n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls =$5C2$ = 10

Therefore, P(E) = n(E)/n(S) = 10/ 21.

Filed Under: Probability

35 24368
Q:

A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a

king of heart is:

 A) 2/13 B) 1/13 C) 1/26 D) 1/52

Explanation:

Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2.

P(E) =n(E)/n(S)=2/52=1/26.

Filed Under: Probability

7 23787
Q:

There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?

 A) 1/2 B) 3/4 C) 4/7 D) 3/8

Explanation:

Total cases of checking in the hotels = 4 x 4 x 4 = 64 ways.

Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways.

Required probability =24/64  = 3/8

Filed Under: Probability

44 23704
Q:

If two letters are taken at random from the word HOME, what is the probability that none of the letters would be vowels?

 A) 1/6 B) 1/2 C) 1/3 D) 1/4

Explanation:

P(first letter is not vowel) = $24$

P(second letter is not vowel) = $13$

So, probability that none of letters would be vowels is = $24×13=16$

Filed Under: Probability

67 22327
Q:

A letter is takenout at random from 'ASSISTANT'  and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

 A) 35/96 B) 19/90 C) 19/96 D) None of these

Explanation:

$ASSISTANT→AAINSSSTT$

$STATISTICS→ACIISSSTTT$

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A =  $2C19C1×1C110C1$ = 1/45

Probability of choosing I = $19C1×2C110C1$ = 1/45

Probability of choosing S = $3C19C1×3C110C1$ = 1/10

Probability of choosing T = $2C19C1×3C110C1$ = 1/15

Hence, Required probability =

Filed Under: Probability
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97 21181
Q:

An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.

 A) 5/204 B) 1/204 C) 13/204 D) None of these

Explanation:

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then

Required probability = $PA∩B∩C$

Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When  a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

$∴PBA=317$

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

Hence the required probability = $29×317×18=1204$

Filed Under: Probability

25 20952
Q:

In a single throw of two dice , find the probability that neither a doublet nor a total of 8 will appear.

 A) 7/15 B) 5/18 C) 13/18 D) 3/16

Explanation:

n(S) = 36

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

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

$∴$Required probability = $PA∪B$

= $PA+PB-PA∩B$

=  $636+536-136$ = $518$

Filed Under: Probability

43 19918
Q:

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

 A) 3/4 B) 3/8 C) 5/16 D) 2/7

Explanation:

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

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

n(E) = 27.

P(E) = n(E)/n(S) = 27/36 = 3/4.