**Question 1 :**

A box contains cards numbered 3, 5, 7, 9, … 35, 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.

**Solution :**

Cards in the box = {3, 5, 7, 9, ............37}

number of cards = n = [(l-a)/d] + 1

n = [(37 - 3)/2] + 1

n = (34/2) + 1

n = 18

n(S) = 18

Let "A" be the event of selecting a number which is multiple of 7

A = {7, 14, 21, 28, 35}

n(A) = 5

P(A) = n(A)/n(S)

P(A) = 5/18

Let "B" be the even of selecting a prime number

B = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37}

n(B) = 11

P(B) = n(B)/n(S)

P(B) = 11/18

A n B = {7}

n(A n B) = 1

P(A n B) = n (A n B)/n(S)

P(A n B) = 1/18

P(A U B) = P(A) + P(B) - P(AnB)

P(A U B) = (5/18) + (11/18) - (1/18)

P(A U B) = (5 + 11 - 1)/18

P(A U B) = 15/18

P(A U B) = 5/6

**Question 2 :**

Three unbiased coins are tossed once. Find the probability of getting atmost 2 tails or atleast 2 heads.

**Solution :**

Sample space = {HHH. HHT, HTH, HTT, THH, THT, TTH, TTT}

n(S) = 8

Let "A" be the event of getting atmost 2 tails

atmost 2 tails = o tail, 1 tail, 2 tails

A = {HHH, HHT, HTH, HTT, THH, THT, TTH}

n(A) = 7

P(A) = n(A)/n(S)

P(A) = 7/8

Let "B" be the event of getting atleast 2 heads.

atleast 2 heads = 2 heads , 3 heads

B = {HHT, HTH, THH, HHH}

n(B) = 4

P(B) = n(B)/n(S)

P(B) = 4/8

A n B = {HHT, HTH,THH, HHH}

P(A n B) = 4/8

P(A U B) = P(A) + P(B) - P(AnB)

P(A U B) = (7/8) + (4/8) - (3/8)

P(A U B) = (7 + 4 - 4)/8

P(A U B) = 7/8

**Question 3 :**

The probability that a person will get an electrification contract is 3/5 and the probability that he will not get plumbing contract is 5/8 . The probability of getting atleast one contract is 5/7. What is the probability that he will get both?

**Solution :**

Let "A" and "B" be the event of getting electrification contract, plumbing contract respectively.

P(A) = 3/5

P(B bar) = 5/8

P(B) = 1 - P(B bar)

= 1 - (5/8)

P(B) = 3/8

P(A u B) = 5/7, we need to find P(A n B) = ?

P(A n B) = P(A) + P(B) - P(A U B)

= (3/5) + (3/8) - (5/7)

= (168 + 105 - 200)/ 280

P(A n B) = 73/280

**Question 4 :**

In a town of 8000 people, 1300 are over 50 years and 3000 are females. It is known that 30% of the females are over 50 years. What is the probability that a chosen individual from the town is either a female or over 50 years?

**Solution :**

Total number of people = 8000

Let "A" and "B" be the event of selecting an individual as female and over 50 years respectively.

Number of people who are over 50 years n(A) = 1300

P(A) = 1300/8000

Number of females n(B) = 3000

P(B) = 3000/8000

n(A n B) = 30% of 3000

= (30/100) 3000

n(A n B) = 900

P(A n B) = 900/8000

P(A U B) = P(A) + P(B) - P(A n B)

= (1300/8000) + (3000/8000) - (900/8000)

= (1300 + 3000 - 900) / 8000

P(A U B) = 3400/8000

P(A U B) = 17/40

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