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**Probability**

### 5 Questions Your Maths Teacher Can Ask To Confuse You

**Q1: A fair coin is tossed twice, find the probability of getting different outcomes of this experiment. What will be the sum of all these probabilities?**

Answer: A coin is tossed twice, the sample space (possible outcomes) will be:

S = {HH, HT, TH, TT}

Let A be the event getting two heads then

Number of Favourable Outcomes for event A |
1 | ||

P(A) = | = | ||

Total Number of Outcomes |
4 |

Let B be the event getting first head and then tail (HT) then

Number of Favourable Outcomes for event B |
1 | ||

P(B) = | = | ||

Total Number of Outcomes |
4 |

Let C be the event getting first tail then head (TH) then

Number of Favourable Outcomes for event C |
1 | ||

P(C) = | = | ||

Total Number of Outcomes |
4 |

Let D be the event getting two tails (TT) then

Number of Favourable Outcomes for event D |
1 | ||

P(D) = | = | ||

Total Number of Outcomes |
4 |

Sum of all probabilities P(A) + P(B) + P(C) + P(D) = 4/4 = 1

**Q2: Find the probability that a leap year selected randomly will have 53 Sundays?**

(

**: If first two days lands on a Sunday in a leap year it would have 53 Sundays. How to prove?).**

*Note*Answer: No. of days in a leap year = 366 days = 52 weeks + 2 days

It implies a leap year will have 52 Sundays.

In remaining 2 days, possible out comes are:

- Sun, Mon

- Mon, Tue

- Tue, Wed

- Wed, Thu

- Thu, Fri

- Fri, Sat

- Sat, Sun

Total out comes = 7

Favourable outcomes that Sunday will come in these two days = 2

∴ Required probability = 2/7

**Q3: A shop keeper has a box containing 10 dolls, out which three are defective. Radha buys a toy from the shop keeper. What is the probability that Radha gets**

**(i) a defective doll.**

**(ii) a non-defective doll.**

Answer: Total Dolls = 10

Defectives = 3

Non-Defective = 7

(i) Probability Radha gets a defective doll

Number of Favourable Outcomes (defective dolls) |
3 | ||

P(A) = | = | ||

Total Number of Outcomes |
10 |

P(A) = 0.3

(ii) Probability Radha gets non-defctive doll

Number of Favourable Outcomes (non-defective dolls) |
7 | ||

P(B) = | = | ||

Total Number of Outcomes |
10 |

P(B) = 0.7

**Q4: A pair of fair dice are rolled. Find the probability that the two dice show the same number.**

Answer: Total possible outcomes when two dice are thrown = 36 = {(1,1), (1,2), ..., (6,5), (6,6)}

Favourable outcomes (two dice show same number) = 6 = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

Number of Favourable Outcomes (two dice show same number) |
6 | ||

P(E) = | = | ||

Total Number of Outcomes |
36 |

P(E) = 1/6

**Q5: In the above example, what will be the probability of getting two different numbers when the two dice are thrown?**

Answer: As computed above, P(A) = getting same numbers on the two dice = 1/6.

∴ Probability Getting two different numbers P(B) = P (NOT A) = 1 - P(A) = 1 - 1/6 = 5/6

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