# 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.** $P(\varnothing )=0$

**iv.** For any events A and B we have :

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

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

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 =$7{C}_{2}$ = 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 =$5{C}_{2}$ = 10

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

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.

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

A) 1/6 | B) 1/2 |

C) 1/3 | D) 1/4 |

Explanation:

P(first letter is not vowel) = $\frac{2}{4}$

P(second letter is not vowel) = $\frac{1}{3}$

So, probability that none of letters would be vowels is = $\frac{2}{4}\times \frac{1}{3}=\frac{1}{6}$

A) 35/96 | B) 19/90 |

C) 19/96 | D) None of these |

Explanation:

$ASSISTANT\to AAINSSSTT$

$STATISTICS\to ACIISSSTTT$

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

Probability of choosing A = $\frac{{2}_{{C}_{1}}}{{9}_{{C}_{1}}}\times \frac{{1}_{{C}_{1}}}{{10}_{{C}_{1}}}$ = 1/45

Probability of choosing I = $\frac{1}{{9}_{{C}_{1}}}\times \frac{{2}_{{C}_{1}}}{{10}_{{C}_{1}}}$ = 1/45

Probability of choosing S = $\frac{{3}_{{C}_{1}}}{{9}_{{C}_{1}}}\times \frac{{3}_{{C}_{1}}}{{10}_{{C}_{1}}}$ = 1/10

Probability of choosing T = $\frac{{2}_{{C}_{1}}}{{9}_{{C}_{1}}}\times \frac{{3}_{{C}_{1}}}{{10}_{{C}_{1}}}$ = 1/15

Hence, Required probability = $\frac{1}{45}+\frac{1}{45}+\frac{1}{10}+\frac{1}{15}=\frac{19}{90}$

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 = $P\left(A\cap B\cap C\right)$

= $P\left(A\right)P\left(\frac{B}{A}\right)P\left(\frac{C}{A\cap B}\right)$

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

$\therefore P\left(\frac{B}{A}\right)=\frac{3}{17}$

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.

$\therefore P\left(\frac{C}{A\cap B}\right)=\frac{2}{16}=\frac{1}{8}$

Hence the required probability = $\frac{2}{9}\times \frac{3}{17}\times \frac{1}{8}=\frac{1}{204}$

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) }

$n\left(A\right)=6,n\left(B\right)=5,n\left(A\cap B\right)=1$

$\therefore $Required probability = $P\left(A\cup B\right)$

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

= $\frac{6}{36}+\frac{5}{36}-\frac{1}{36}$ = $\frac{5}{18}$

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.