# 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:

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

Explanation:

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

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

Filed Under: Probability

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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

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.

Filed Under: Probability

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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

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.

$4C2$

=>$4*32*1$

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

Filed Under: Probability

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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

Explanation:

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

Number of white balls = 8.

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

Filed Under: Probability

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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

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.

Filed Under: Probability

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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

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.

Filed Under: Probability

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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

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.

Filed Under: Probability

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Q:

Three unbiased coins are tossed.What is the probability of getting at least 2 heads?

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

Explanation:

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

Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.

P(E) = n(E) / n(S)

= 4/8= 1/2