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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