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Probability

Short Answer Type

981. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var(X).

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Long Answer Type

982. Find the variance of the number obtained on a throw of an unbiased die.

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983. A die is tossed twice. Getting a number greater than 4 is considered a success. Find the variance of the probability distribution of the number of successes.

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984. Two bad eggs are mixed accidentally with 10 good ones. Three eggs are drawn at random without replacement from this lot. Find the mean and variance for the number of bad eggs.

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985. Three bad eggs got mixed up with 7 good eggs. If three eggs are drawn (without replacement) from 10 eggs, find the mean and variance for the number of bad eggs among them.

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986. Five defective bulbs are accidentally mixed with twenty good ones. It is not possible to just look at a bulb and tell whether or not a bulb is defective. Four bulbs are drawn at random from this lot. Find the mean number of defective bulbs drawn.

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987. Two cards are drawn simultaneously (or successively, without replacement) from a well-shuffled deck of 52 cards. Compute σ² for the number of aces.

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988. In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.

If X is the number of throws, then X may take values 1, 2, 3.
P(X = 1) = P(man gets a six on first throw) =

1 over 6

P(X = 2) = P(man gets a six on first throw but gets a six on second throw)

equals space open parentheses 1 minus 1 over 6 close parentheses space open parentheses 1 over 6 close parentheses space equals space 5 over 6 space cross times space 1 over 6 space equals space 5 over 36

P(X = 3) = P(man does not get a six on first and second throws)

equals space open parentheses 1 minus 1 over 6 close parentheses space open parentheses 1 minus 1 over 6 close parentheses space equals space 5 over 6 cross times 5 over 6 space equals space 25 over 36

∴ probability distribution of X is

When X = 1, the man gains Re 1.
When X = 2, the man does not gain anything.
[∵ on first throw he loses Re 1 and on second throw he gains Re 1]
When X = 3,
(i) the man may lose Re 3 when all the three throws show a non-six, which happens with probability $5 over 6 cross times 5 over 6 cross times 5 over 6 space equals space 125 over 216.$
(ii) the man may lose Re 1 when first two throws show a non-six and third shows a six, which happens with probability $5 over 6 cross times 5 over 6 cross times 1 over 6 space equals space 25 over 216.$
If Y is the amount gained or lost, then Y takes values 1, 0, - 3, - 1.
∴ probability distribution of Y is

Expected value of Y = $1 cross times 1 over 6 plus 0 cross times 5 over 36 plus left parenthesis negative 3 right parenthesis space cross times 125 over 216 plus left parenthesis negative 1 right parenthesis space cross times space 25 over 216$

$equals space 1 over 6 plus 0 minus 375 over 216 minus 25 over 216 equals space fraction numerator 36 minus 375 minus 25 over denominator 216 end fraction space equals space minus 364 over 216 space equals space minus 91 over 54$
Hence, the man is expected to lose Rs. $91 over 54$