SOLUTION: 1) There are 10 tennis balls in a box; 6 of them are new. Then two balls are taken from the box for the first play and are not returned there after the play. Then two random balls
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-> SOLUTION: 1) There are 10 tennis balls in a box; 6 of them are new. Then two balls are taken from the box for the first play and are not returned there after the play. Then two random balls
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Question 1204234: 1) There are 10 tennis balls in a box; 6 of them are new. Then two balls are taken from the box for the first play and are not returned there after the play. Then two random balls are taken for the second play from this box. Compute the probability that two balls taken for the second play are new.
2) There are eleven cars in some company. Four of them have problems with electronics. If five cars are randomly selected for inspection, what is the probability three of them do not have problems with electronics? Answer by ikleyn(52787) (Show Source):
You can put this solution on YOUR website! .
1) There are 10 tennis balls in a box; 6 of them are new. Then two balls are taken from the box
for the first play and are not returned there after the play. Then two random balls are taken
for the second play from this box. Compute the probability that two balls taken for the second play are new.
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I will solve here only one, FIRST problem.
The probability that first two balls are new is P = = .
The probability that first two balls are old is P = = .
The probability that of the first two balls one is new, while the other is old is the complement, i.e. .
After taking first two balls, we have the remaining content in the box
8 = 10-2 = 4 new + 4 old with the probability ;
8 = 10-2 = 6 new + 2 old with the probability ;
8 = 10-2 = 5 new + 3 old with the probability .
Now the probability that two balls taken for the second play are new is this weighted sum
P = = = = . ANSWER
Solved.
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