# 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) 1/2 | B) 7/15 |

C) 8/15 | D) 1/9 |

Explanation:

Let S be the sample space

Then n(S) = no of ways of drawing 2 balls out of (6+4) =$10{C}_{2}$ 10 =$\frac{10*9}{2*1}$ =45

Let E = event of getting both balls of same colour

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

=$6{C}_{2}+4{C}_{2}$ = $\frac{6*5}{2*1}+\frac{4*3}{2*1}$= 15+6 = 21

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

A) 1/4 | B) 1/2 |

C) 3/4 | D) 7/12 |

Explanation:

Let A, B, C be the respective events of solving the problem and $\overline{)A},\overline{)B},\overline{)C}$ be the respective events of not solving the problem. Then A, B, C are independent event

$\therefore \overline{)A},\overline{)B},\overline{)C}$ are independent events

Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

$P\left(\overline{)A}\right)=\frac{1}{2},P\left(\overline{)B}\right)=\frac{2}{3},P\left(\overline{)C}\right)=\frac{3}{4}$

$\therefore $ P( none solves the problem) = P(not A) and (not B) and (not C)

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

= $P\left(\overline{)A}\right)P\left(\overline{)B}\right)P\left(\overline{)C}\right)$ $\left[\because \overline{)A},\overline{)B},\overline{)C}areIndependent\right]$

= $\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}$

= $\frac{1}{4}$

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= $1-\frac{1}{4}$= **3/4**

A) 52/221 | B) 55/190 |

C) 55/221 | D) 19/221 |

Explanation:

We have n(s) =$52{C}_{2}$ 52 = 52*51/2*1= 1326.

Let A = event of getting both black cards

B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = $\frac{52*51}{2*1}$ = $26{C}_{2}$ = 325, n(B)= $\frac{26*25}{2*1}$= 4*3/2*1= 6 and n(A∩B) = $4{C}_{2}$ = 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

A) 1/2 | B) 3/5 |

C) 9/20 | D) 8/15 |

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

A) 5/12 | B) 1/6 |

C) 1/2 | D) 7/9 |

Explanation:

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

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

n(E) = 15.

P(E) = n(E)/n(S) = 15/36 = 5/12.

A) 2/91 | B) 1/22 |

C) 3/22 | D) 2/77 |

Explanation:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = $15{C}_{3}$ =$\frac{15*14*13}{3*2*1}$= 455.

Let E = event of getting all the 3 red balls.

n(E) = $5{C}_{3}$ = $\frac{5*4}{2*1}$ = 10.

=> P(E) = n(E)/n(S) = 10/455 = 2/91.

A) 2/7 | B) 1/7 |

C) 3/4 | D) 4/5 |

A) 2/7 | B) 5/7 |

C) 1/5 | D) 1/2 |

Explanation:

Total number of outcomes possible, n(S) = 10 + 25 = 35

Total number of prizes, n(E) = 10

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{10}{35}=\frac{2}{7}$