# Probability Questions

**FACTS AND FORMULAE FOR PROBABILITY QUESTIONS**

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**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) 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.

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.

A) 1/15 | B) 1/221 |

C) 25/57 | D) 35/256 |

Explanation:

Let S be the sample space

Then, n(S) = $52{C}_{2}$= 1326.

Let E = event of getting 2 kings out of 4.

n(E) = $\frac{52*51}{2*1}$= 6.

$4{C}_{2}$

=>$\frac{4*3}{2*1}$

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{6}{1326}=\frac{1}{221}$

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.

A) 3/20 | B) 29/34 |

C) 47/100 | D) 13/102 |

Explanation:

Let S be the sample space.

Then, n(S) = $52{C}_{2}$=(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 = $13{C}_{1}*13{C}_{1}$ = 169.

P(E) = n(E)/n(S) = 169/1326 = 13/102.

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.

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.

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