The probability distribution of discrete r . v . X is as follows X = x 1 2 3 4 5 6 P X = x k 2 k 3 k 4 k 5 k 6 k Determine the value of k .

Given that X = x 1 2 3 4 5 6 P X = x k 2 k 3 k 4 k 5 k 6 k We know that in probability distribution table sum of all the probability is equal to 1 . So, P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 ) + P ( X = 5 ) + P ( X = 6 ) = 1 ⇒ k + 2 k + 3 k + 4 k + 5 k + 6 k = 1 ⇒ K = 1 21

The probability distribution of discrete r . v . X is as follows X = x 1 2 3 4 5 6 P X = x k 2 k 3 k 4 k 5 k 6 k Find P X ≤ 4 , P 2 < X < 4 , P X ≥ 3 .

Given that X = x 1 2 3 4 5 6 P X = x k 2 k 3 k 4 k 5 k 6 k We know that in probability distribution table sum of all the probability is equal to 1 . So, P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 ) + P ( X = 5 ) + P ( X = 6 ) = 1 ⇒ k + 2 k + 3 k + 4 k + 5 k + 6 k = 1 ⇒ K = 1 21 (i) P ( X ≤ 4 ) = P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 ) = k + 2 k + 3 k + 4 k = 10 k = 10 21 (ii) P ( 2 < X < 4 ) = P ( X = 3 ) = 3 k = 3 21 = 1 7 (iii) P ( X ≥ 3 ) = P ( X = 3 ) + P ( X = 4 ) + P ( X = 5 ) + P ( X = 6 ) = 3 k + 4 k + 5 k + 6 k = 18 k = 18 21 = 6 7

From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs are drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Out of 30 bulbs, 6 are defective. ⇒ Number of non-defective bulbs = 30 - 6 = 24 Probability of drawing a defective bulb = 6 30 = 1 5 Probability of drawing a non-defective bulb = 24 30 = 4 5 4 bulbs are drawn from the lot with replacement. Let X be the random variable that denotes the number of defective bulbs in the selected bulbs. ∴ P X = 0 = P ( 4 non-defective and 0 defective) = C 0 4 × 4 5 4 × 1 5 0 = 256 625 P X = 1 = P ( 3 non-defective and 1 defective) = C 1 4 × 1 5 × 4 5 3 = 256 625 P X = 2 = P ( 2 non-defective and 2 defective) = C 2 4 × 1 5 2 × 4 5 2 = 96 625 P X = 3 = P ( 1 non-defective and 3 defective) = C 3 4 × 1 5 3 × 4 5 = 16 625 P X = 4 = P ( 0 non-defective and 4 defective) = C 4 4 × 1 5 4 × 4 5 0 = 1 625 Therefore, the required probability distribution is as follows: X 0 1 2 3 4 P X 256 625 256 625 96 625 16 625 1 625

A random variable X has the following probability distribution: X 0 1 2 3 4 5 6 7 P X 0 k 2 k 2 k 3 k k 2 2 k 2 7 k 2 + k Find k .

Given that, X 0 1 2 3 4 5 6 7 P X 0 k 2 k 2 k 3 k k 2 2 k 2 7 k 2 + k Since the sum of all the probability in a probability distribution is always unity. ∴ P ( X = 0 ) + P ( x = 1 ) + . . . . . . + P ( x = 7 ) = 1 = 0 + k + 2 k + 2 k + 3 k + k 2 + 2 k 2 + 7 k 2 + k = 1 = 10 k 2 + 9 k - 1 = 0 ⇒ ( 10 k - 1 ) ( k + 1 ) = 0 ⇒ k ≠ - 1 ∵ k ≥ 0 ⇒ 10 k - 1 = 0 ⇒ k = 1 10

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If the $a . d . r . v . X$ has the following probability distribution:

$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P (X = x)$ $k$ $2 k$ $2 K$ $3 k$ $k^{2}$ $2 k^{2}$ $7 k^{2} + k$

then $k =$

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Important Questions on Probability Distributions

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Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 7 - Probability Distributions>MISCELLANEOUS EXERCISE 7>Q 10

The expected value of $X$ for the following $p . m . f .$

$X$	$- 2$	$- 1$	$0$	$1$	$2$
$P (X)$	$0.3$	$0.4$	$0.2$	$0.15$	$0.25$

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12th Maharashtra Board

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Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 7 - Probability Distributions>MISCELLANEOUS EXERCISE 7>Q 1

Identify the random variable as either discrete or continuous in the following. If the random variable is discrete, list its possible values.

An economist is interested in the random variable $X,$ the number of unemployed graduates in a town of population $1$ lakh.

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12th Maharashtra Board

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Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 7 - Probability Distributions>MISCELLANEOUS EXERCISE 7>Q 1

Identify the random variable as either discrete or continuous in each of the following. If the random variable is discrete, list its possible values.

The random variable $X$ is the amount of syrup prescribed by a physician

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12th Maharashtra Board

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Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 7 - Probability Distributions>MISCELLANEOUS EXERCISE 7>Q 1

Identify the random variable as either discrete or continuous in each of the following. If the random variable is discrete, list its possible values.

A person on a high protein diet is interested in the random variable $X,$ the gain in weight in a week.

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12th Maharashtra Board

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Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 7 - Probability Distributions>MISCELLANEOUS EXERCISE 7>Q 1

Identify the random variable as either discrete or continuous in each of the following. Write down the range of it.

$20$ white rats are available for an experiment. Twelve rats are male. Scientist randomly selects $5$ rats number of female rats selected on a specific day.

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12th Maharashtra Board

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Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 7 - Probability Distributions>MISCELLANEOUS EXERCISE 7>Q 1

Identify the random variable as either discrete or continuous in each of the following. If the random variable is discrete, list its possible values.

A highway-safety group is interested in studying the speed (in $km / h$ ) of a car at a checkpoint.

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12th Maharashtra Board

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Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 7 - Probability Distributions>MISCELLANEOUS EXERCISE 7>Q 2

The probability distribution of discrete $r . v . X$ is as follows

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P (X = x)$	$k$	$2 k$	$3 k$	$4 k$	$5 k$	$6 k$

Determine the value of $k$ .

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Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 7 - Probability Distributions>MISCELLANEOUS EXERCISE 7>Q 2

The probability distribution of discrete $r . v . X$ is as follows

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P (X = x)$	$k$	$2 k$	$3 k$	$4 k$	$5 k$	$6 k$

Find $P (X \leq 4), P (2 < X < 4), P (X \geq 3)$ .

If the a.d.r.v. X has the following probability distribution: X 1 2 3 4 5 6 7 PX=x k 2k 2K 3k k2 2k2 7k2+k then k=

Important Questions on Probability Distributions

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If the $a . d . r . v . X$ has the following probability distribution:

$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P (X = x)$ $k$ $2 k$ $2 K$ $3 k$ $k^{2}$ $2 k^{2}$ $7 k^{2} + k$

then $k =$