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Probability

Short Answer Type

1061. An unbiased coin is tossed six times. Find the probability of obtaining at least 2 heads.

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1062. In a box containing 100 bulbs, 10 are defective. What is the probabilities that out of a sample of 5 bulbs
(i) none is defective (ii) exactly two are defective.

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1063. If “getting a 6” in a throw of an unbiased die is a success and the random variable x denotes the number of successes in five throws of the die, find P (2 < x < 5)

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1064. If “getting a 5 or a 6” in the throw of an unbiased die is a success and the random variable x denotes the number of successes in 6 throws of the die, find P (x ≤ 4).

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Long Answer Type

1065. A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the dice.

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Short Answer Type

1066. Eight coins are thrown simultaneously. Find the probability of getting at least six heads.

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Long Answer Type

1067.

A pair of dice is thrown 7 times. If getting a total of 7 is considered a success, what is the probability of
(i) no success? (ii) 6 successes? (iii) at least 6 successes? (iv) at most 6 successes?

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1068. Suppose X has a binomial distribution B

open parentheses 6 comma space 1 half close parentheses.

Show that X = 3 is the most likely outcome.

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1069. Find the mean of the Binomial distribution $straight B space open parentheses 4 comma space 1 third close parentheses.$

Let X be the random variable whose probability distribution is $straight B open parentheses 4 comma space 1 third close parentheses.$
$therefore space space straight n space equals space 4 comma space space space straight p space equals space 1 third space space and space straight q space equals space 1 minus 1 third space equals 2 over 3.$
Now $straight P left parenthesis straight X space equals straight x right parenthesis space equals space straight C presuperscript 4 subscript straight x space open parentheses 2 over 3 close parentheses to the power of 4 minus straight x end exponent space open parentheses 1 third close parentheses to the power of straight x comma space space space straight x space equals space 0 comma space 1 comma space 2 comma space 3 comma space 4.$
∴ the distribution of X is

$therefore$ mean = $sum from straight i space equals space 1 to 4 of straight x subscript straight i space straight p left parenthesis straight x subscript straight i right parenthesis space$
$equals space 0 space plus space straight C presuperscript 4 subscript 1 space open parentheses 2 over 3 close parentheses cubed space open parentheses 1 third close parentheses plus 2. space straight C presuperscript 4 subscript 2 space open parentheses 2 over 3 close parentheses squared space open parentheses 1 third close parentheses squared$
$plus 3. space space straight C presuperscript 4 subscript 3 space open parentheses 2 over 3 close parentheses space open parentheses 1 third close parentheses cubed plus 4 space straight C presuperscript 4 subscript 4 space open parentheses 1 third close parentheses to the power of 4$
$equals space 4 cross times space 2 cubed over 3 to the power of 4 plus 2 space cross times space fraction numerator 4 space cross times space 3 over denominator 1 space cross times 2 end fraction cross times space 2 squared over 3 to the power of 4 plus 3 space cross times space 4 over 1 space cross times space 2 over 3 to the power of 4 plus 4 space cross times space 1 space cross times space 1 over 3 to the power of 4 equals space fraction numerator 32 plus 48 plus 24 plus 4 over denominator 3 to the power of 4 end fraction equals space 108 over 81 equals space 4 over 3.$