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91. In a race where 12 cars are running, the chance that car X will win is 16, that Y will win is 110 and that Z will win is 18. Assuming that a dead heat is impossible. Find the chance that one of them will win.

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Solution:
Required probability = P(X) + P(Y) + P(Z) (all the events are mutually exclusive)
=16+110+18=47120
92. A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

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Solution:
P(odd)=P(even)=12     (because there are 50 odd and 50 even numbers)
Sum or the three numbers can be odd only under the following 4 scenarios:
odd+odd+odd=12×12×12=18odd+even+even=12×12×12=18even+odd+even=12×12×12=18even+even+odd=12×12×12=18
Other combinations of odd and even will give even numbers.
Adding up the 4 scenarios above:
=18+18+18+18=48=12
93. A special lottery is to be held to select a student who will live in the only deluxe room in a hostel. There are 100 Year-III, 150 Year-II and 200 Year-I students who applied.
Each Year-III's name is placed in the lottery 3 times; each Year-II's name, 2 times and Year-I's name, 1 time. What is the probability that a Year-III's name will be chosen?

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Solution:
Total names in the lottery,
=3×100+2×150+200=800
Number of Year-III's names,
=3×100=300
Required probability,
=300800=38
94. From a bag containing 4 white and 5 black balls a man drawn 3 balls at random. What are the odds against these balls being black?

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Solution:
Probability of all three balls being black
=5C39C3=542
Probability that three balls are not black
=1542=3742
Hence, odds against these ball being black
=(3742):(542)=37:5
95. A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag.
Find the probability that one ball is red and one is green.

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Solution:
Let A be the event that ball selected from the first bag is red and ball selected from second bag is green.
Let B be the event that ball selected from the first bag is green and ball selected from second bag is red.
P(A)=(58)×(610)=38andP(B)=(38)×(410)=320
Hence, required probability,
=P(A)+P(B)=38+320=2140
96. The probability of a lottery ticket being a prized ticket is 0.2. When 4 tickets are purchased, the probability of winning a prize on at least one ticket is -

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Solution:
P(winning prize at least on one ticket)
= 1 - P("Losing on all tickets")
= 1 - (0.8)4 = (1 + (0.8)2)(1 - (0.8)2)
= (1.64)(0.36)
= 0.5904
97. A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that all the four bulbs are defective.

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Solution:
Out of nine, five are good and four are defective.
Required probability
=4C49C4=1126
98. If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is -

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Solution:
The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other.
Let the event of getting either both even or both odd
Then
P(E¯)=1836=12P(E)=112=12