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Probability

Long Answer Type

1071. If a fair coin is tossed 10 times, find the probability of
(i) exactly six heads
(ii) at least six heads
(iii) at most six heads

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1072. A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is

1 over 100.

What is the probaility that he will win a prize exactly once (a) at least once (b) exactly once (c) at least twice?

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1073.

An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark ‘Y’ . A ball is drawn at random and it is replaced. If 6 balls are drawn in this way, find the probability that
(i) all will bear mark X
(ii) not more than 2 balls will bear 'Y' mark.
(iii) at least one ball will bear 'Y' mark.
(iv) the number of balls with 'X' mark and 'Y' mark will be equal.

104 Views

Short Answer Type

1074. In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true’ ; if it falls tails, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.

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1075. Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

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1076. How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

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Multiple Choice Questions

1077. In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

$10 to the power of negative 1 end exponent$
$open parentheses 1 half close parentheses to the power of 5$
$open parentheses 9 over 10 close parentheses to the power of 5$
$open parentheses 9 over 10 close parentheses to the power of 5$

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1078. The probability that a student is not a swimmer is

1 fifth.

Then the probability that out of five students, four are swimmers is

$straight C presuperscript 5 subscript 4 space open parentheses 4 over 5 close parentheses to the power of 4 space 1 fifth$
$open parentheses 4 over 5 close parentheses to the power of 4 space 1 fifth$
$straight C presuperscript 5 subscript 1 space open parentheses 4 over 5 close parentheses to the power of 4$
$straight C presuperscript 5 subscript 1 space open parentheses 4 over 5 close parentheses to the power of 4$

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Long Answer Type

1079. A and B throw two dice simultaneously turn by turn. A will win if he throws a total of 5, B will win if he throws a doublet. Find the probability that B will win the game, though A started the game.

∴ total number of possible outcomes = 36
Let E be the event of getting "a total of 5" in a throw of two dice.

therefore space space space space straight E space equals space left curly bracket left parenthesis 1 comma space 4 right parenthesis comma space left parenthesis 2 comma space 3 right parenthesis comma space left parenthesis 3 comma space 2 right parenthesis comma space left parenthesis 4 comma space 1 right parenthesis right curly bracket therefore space space space space straight P left parenthesis straight E right parenthesis space equals space 4 over 36 space equals space 1 over 9 comma space space space straight P left parenthesis straight E with bar on top right parenthesis space equals space 1 minus 1 over 9 space equals space 8 over 9

Let F be the event of getting a doublet in throw of two dice.

therefore

F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

therefore space space space straight P left parenthesis straight F right parenthesis space equals space 6 over 36 space equals space 1 over 6 comma space space straight P left parenthesis straight F with bar on top right parenthesis space equals space 1 minus 1 over 6 space equals 5 over 6

Now A starts the game.

therefore

B wins in second throw or fourth throw or 6th throw or ..........

therefore

probability of winning of B =

straight P left parenthesis straight E with bar on top right parenthesis space space straight P left parenthesis straight F right parenthesis space plus space straight P left parenthesis straight E with bar on top right parenthesis thin space straight P left parenthesis straight F with bar on top right parenthesis thin space straight P left parenthesis straight E with bar on top right parenthesis space straight P left parenthesis straight F right parenthesis

plus space straight P left parenthesis straight E with bar on top right parenthesis thin space straight P left parenthesis straight F with bar on top right parenthesis thin space straight P left parenthesis straight E with bar on top right parenthesis space straight P left parenthesis straight F with bar on top right parenthesis thin space straight P left parenthesis straight E with bar on top right parenthesis thin space straight P left parenthesis straight F right parenthesis space plus.......

equals space 8 over 9 cross times 1 over 6 plus 8 over 9 cross times 5 over 6 cross times 8 over 9 cross times 1 over 6 plus 8 over 9 cross times 5 over 6 cross times 8 over 9 cross times 5 over 6 cross times 8 over 9 cross times 1 over 6 plus..... equals space 8 over 9 cross times 1 over 6 open square brackets 1 plus space open parentheses 5 over 6 cross times 8 over 9 close parentheses space plus space open parentheses 5 over 6 cross times 8 over 9 close parentheses squared plus..... close square brackets equals space 8 over 9 cross times space 1 over 6 open square brackets fraction numerator 1 over denominator 1 minus begin display style 5 over 6 end style cross times begin display style 8 over 9 end style end fraction close square brackets space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space straight S space equals space fraction numerator straight a over denominator 1 minus straight r end fraction close square brackets equals space 8 over 9 cross times 1 over 6 open square brackets fraction numerator 6 space cross times space 9 over denominator 6 space cross times 9 space minus space 5 cross times 8 end fraction close square brackets space equals space 8 over 9 cross times 1 over 6 open square brackets fraction numerator 6 space cross times space 9 over denominator 14 end fraction close square brackets space equals space 8 over 9 cross times space 1 over 6 cross times fraction numerator 6 space cross times space 9 over denominator 14 end fraction equals space 4 over 7

176 Views

Short Answer Type

1080.

A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y.

1714 Views