There are 2 different blue toys, 3 different green toys and 4 different red toys. In how many ways one can choose 3 toys, such that there is at least one green toy in the chosen three?
Sample can be (1 green + 2 non-green) or (2 green + 1 non-green) or (3 green).
Number of ways = (3C1 * 6C2) + (3C2 * 6C1) + (3C3)= 45+18+1= 64 ways.
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Among a group of 2500 people, 35 percent invest in municipal bonds, 18 percent invest in oil stocks, and 7 percent invest in both municipal bonds and oil stocks. If 1 person is to be randomly selected from 2500 people, what is the probability that the person selected will be one who invests in municipal bonds but not in oil stocks?
Probability of hitting the target = (1 - probability of not hitting the target)
if man hits once every 3 balls,
probability of hitting = 1/3 => probability of not hitting=2/3
So, probability of not hitting the target all three times = (2/3)* (2/3)* (2/3) = 8/27
=> Probability of hitting target = (1 - 8/27) = 19/27
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4 men throw a die each simultaneously. Find the probability that at least 2 people get the same number
A leap year has 366 day which is 52 full weeks + 2 odd days. Now these two odd days may be (sun + mon), (mon + tue), .... (Sat + sun). Now there are total 7 ways. Of which Sunday appeared two times. So answer 2/7
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At a car park there are 100 vehicles, 60 of which are cars, 30 are vans and the remaining are Lorries. If every vehicle is equally likely to leave, find the probability of car leaving second if either a lorry or van had left first :
Let S be the sample space and A be the event of a van leaving first. n(S) = 100 n (A) = 30
Probability of a van leaving first: P (A) = 30/100 = 3/10
Let B be the event of a lorry leaving first. n (B) = 100 - 60 - 30 = 10 Probability of a lorry leaving first: P (B) = 10/100 = 1/10
If either a lorry or van had left first, then there would be 99 vehicles remaining, 60 of which are cars.
Let T be the sample space and C be the event of a car leaving. n (T) = 99 n(C) = 60
Probability of a car leaving after a lorry or van has left: P(C) = 60/99 = 20/33
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Three terrorists are employed to shoot a renowned person Mr. X. Only one bullet is sufficient to kill him if it strikes in the head. The probabilities of the terrorists striking Mr. X's head by their bullets are 0.2, 0.3, and 0.4. What is the probability that Mr. X is shot dead?
Probability of Mr. X shot dead = Probability of atleast one terrorist shoot on head
= 1 - (Probability of None shoot on head)
=1-(0.8*0.7*0.6)
=0.664
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There are two bags containing white and black balls. In the first bag there are 8 white and 6 black balls and in the second bag, there are 4 white and 7 black balls. One ball is drawn at random from any of these two bags. Find the probability of this ball being black.
The number of ways of rolling a dice where no two numbers probability that no one rolls the same number = 6*5*4*3
Now total possibilities of rolling a dice = 64
The probability that a no one gets the same number = 6*5*4*3/64=5/18
So the probability that at least two people gets same number = 1?5/18=13/18
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