Advertisements
Advertisements
प्रश्न
Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X.
उत्तर १
The random variable X, i.e., the larger of the two selected numbers, takes the values 3, 4, 5, 6 or 7.
\[P\left( X = 3 \right) = P\left\{ \left( 2, 3 \right), \left( 3, 2 \right) \right\} = 2 \times \frac{1}{6} \times \frac{1}{5} = \frac{2}{30}\]
\[P\left( X = 4 \right) = P\left\{ \left( 2, 4 \right), \left( 3, 4 \right), \left( 4, 2 \right), \left( 4, 3 \right) \right\} = 4 \times \frac{1}{6} \times \frac{1}{5} = \frac{4}{30}\]
\[P\left( X = 5 \right) = P\left\{ \left( 2, 5 \right), \left( 3, 5 \right), \left( 4, 5 \right), \left( 5, 2 \right), \left( 5, 3 \right), \left( 5, 4 \right) \right\} = 6 \times \frac{1}{6} \times \frac{1}{5} = \frac{6}{30}\]
\[P\left( X = 6 \right) = P\left\{ \left( 2, 6 \right), \left( 3, 6 \right), \left( 4, 6 \right), \left( 5, 6 \right), \left( 6, 2 \right), \left( 6, 3 \right), \left( 6, 4 \right), \left( 6, 5 \right) \right\} = 8 \times \frac{1}{6} \times \frac{1}{5} = \frac{8}{30}\]
\[P\left( X = 7 \right) = P\left\{ \left( 2, 7 \right), \left( 3, 7 \right), \left( 4, 7 \right), \left( 5, 7 \right), \left( 6, 7 \right), \left( 7, 2 \right), \left( 7, 3 \right), \left( 7, 4 \right), \left( 7, 5 \right), \left( 7, 6 \right) \right\} = 10 \times \frac{1}{6} \times \frac{1}{5} = \frac{10}{30}\]
Thus, the probability distribution of X is
उत्तर २
The random variable X, i.e., the larger of the two selected numbers, takes the values 3, 4, 5, 6 or 7.
\[P\left( X = 3 \right) = P\left\{ \left( 2, 3 \right), \left( 3, 2 \right) \right\} = 2 \times \frac{1}{6} \times \frac{1}{5} = \frac{2}{30}\]
\[P\left( X = 4 \right) = P\left\{ \left( 2, 4 \right), \left( 3, 4 \right), \left( 4, 2 \right), \left( 4, 3 \right) \right\} = 4 \times \frac{1}{6} \times \frac{1}{5} = \frac{4}{30}\]
\[P\left( X = 5 \right) = P\left\{ \left( 2, 5 \right), \left( 3, 5 \right), \left( 4, 5 \right), \left( 5, 2 \right), \left( 5, 3 \right), \left( 5, 4 \right) \right\} = 6 \times \frac{1}{6} \times \frac{1}{5} = \frac{6}{30}\]
\[P\left( X = 6 \right) = P\left\{ \left( 2, 6 \right), \left( 3, 6 \right), \left( 4, 6 \right), \left( 5, 6 \right), \left( 6, 2 \right), \left( 6, 3 \right), \left( 6, 4 \right), \left( 6, 5 \right) \right\} = 8 \times \frac{1}{6} \times \frac{1}{5} = \frac{8}{30}\]
\[P\left( X = 7 \right) = P\left\{ \left( 2, 7 \right), \left( 3, 7 \right), \left( 4, 7 \right), \left( 5, 7 \right), \left( 6, 7 \right), \left( 7, 2 \right), \left( 7, 3 \right), \left( 7, 4 \right), \left( 7, 5 \right), \left( 7, 6 \right) \right\} = 10 \times \frac{1}{6} \times \frac{1}{5} = \frac{10}{30}\]
Thus, the probability distribution of X is
| X | 3 | 4 | 5 | 6 |
| P(X) | `2/30` | `4/30` | `6/30` | `8/30` |
\[\therefore \text { Mean }, E(X) = \Sigma x_i p_i = 3 \times \frac{2}{30} + 4 \times \frac{4}{30} + 5 \times \frac{6}{30} + 6 \times \frac{8}{30} + 7 \times \frac{10}{30}\]
\[ = \frac{6}{30} + \frac{16}{30} + \frac{30}{30} + \frac{48}{30} + \frac{70}{30}\]
\[ = \frac{170}{30} = \frac{17}{3}\]
\[E( X^2 ) = \Sigma x_i^2 p_i = 3^2 \times \frac{2}{30} + 4^2 \times \frac{4}{30} + 5^2 \times \frac{6}{30} + 6^2 \times \frac{8}{30} + 7^2 \times \frac{10}{30}\]
\[ = \frac{18}{30} + \frac{64}{30} + \frac{150}{30} + \frac{288}{30} + \frac{490}{30}\]
\[ = \frac{1010}{30} = \frac{101}{3}\]
Thus,
\[\text { Var }\left( X \right) = E\left( X^2 \right) - \left[ E\left( X \right) \right]^2\]
\[= \frac{101}{3} - \left( \frac{17}{3} \right)^2 \]
\[ = \frac{303 - 289}{9} = \frac{14}{9}\]
APPEARS IN
संबंधित प्रश्न
From a lot of 15 bulbs which include 5 defectives, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence find the mean of the distribution.
Probability distribution of X is given by
| X = x | 1 | 2 | 3 | 4 |
| P(X = x) | 0.1 | 0.3 | 0.4 | 0.2 |
Find P(X ≥ 2) and obtain cumulative distribution function of X
Find the probability distribution of number of heads in four tosses of a coin.
If the probability that a fluorescent light has a useful life of at least 800 hours is 0.9, find the probabilities that among 20 such lights at least 2 will not have a useful life of at least 800 hours. [Given : (0⋅9)19 = 0⋅1348]
There are 4 cards numbered 1 to 4, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two drawn cards. Find the mean and variance of X.
Let X be a random variable which assumes values x1, x2, x3, x4 such that 2P (X = x1) = 3P(X = x2) = P (X = x3) = 5 P (X = x4). Find the probability distribution of X.
A bag contains 4 red and 6 black balls. Three balls are drawn at random. Find the probability distribution of the number of red balls.
Five defective mangoes are accidently mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.
Two dice are thrown together and the number appearing on them noted. X denotes the sum of the two numbers. Assuming that all the 36 outcomes are equally likely, what is the probability distribution of X?
Let X represent the difference between the number of heads and the number of tails when a coin is tossed 6 times. What are the possible values of X?
If the probability distribution of a random variable X is given by Write the value of k.
| X = xi : | 1 | 2 | 3 | 4 |
| P (X = xi) : | 2k | 4k | 3k | k |
A random variable X has the following probability distribution:
| X : | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P (X) : | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
For the events E = {X : X is a prime number}, F = {X : X < 4}, the probability P (E ∪ F) is
If X is a random-variable with probability distribution as given below:
| X = xi : | 0 | 1 | 2 | 3 |
| P (X = xi) : | k | 3 k | 3 k | k |
The value of k and its variance are
From a lot of 15 bulbs which include 5 defective, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.
A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of number of successes.
Three fair coins are tossed simultaneously. If X denotes the number of heads, find the probability distribution of X.
Amit and Rohit started a business by investing ₹20,000 each. After 3 months Amit withdrew ₹5,000 and Rohit put in ₹5,000 additionally. How should a profit of ₹12,800 be divided between them at the end of the year?
The probability that a bulb produced by a factory will fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5) that fuse after 200 days of use. Find the probability of X ≤ 1
10 balls are marked with digits 0 to 9. If four balls are selected with replacement. What is the probability that none is marked 0?
State whether the following is True or False :
If r.v. X assumes the values 1, 2, 3, ……. 9 with equal probabilities, E(x) = 5.
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as six appears on at least one die
The probability distribution of a random variable X is given below:
| X | 0 | 1 | 2 | 3 |
| P(X) | k | `"k"/2` | `"k"/4` | `"k"/8` |
Find P(X ≤ 2) + P (X > 2)
Let X be a discrete random variable whose probability distribution is defined as follows:
P(X = x) = `{{:("k"(x + 1), "for" x = 1"," 2"," 3"," 4),(2"k"x, "for" x = 5"," 6"," 7),(0, "Otherwise"):}`
where k is a constant. Calculate Standard deviation of X.
The probability distribution of a random variable x is given as under:
P(X = x) = `{{:("k"x^2, "for" x = 1"," 2"," 3),(2"k"x, "for" x = 4"," 5"," 6),(0, "otherwise"):}`
where k is a constant. Calculate E(3X2)
A random variable x has to following probability distribution.
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Determine
A person throws two fair dice. He wins ₹ 15 for throwing a doublet (same numbers on the two dice), wins ₹ 12 when the throw results in the sum of 9, and loses ₹ 6 for any other outcome on the throw. Then the expected gain/loss (in ₹) of the person is ______.
A large chain retailer purchases an electric device from the manufacturer. The manufacturer indicates that the defective rate of the device is 10%. The inspector of the retailer randomly selects 4 items from a shipment. Complete the following activity to find the probability that the inspector finds at most one defective item in the 4 selected items.
Solution:
Here, n = 4
p = probability of defective device = 10% = `10/100 = square`
∴ q = 1 - p = 1 - 0.1 = `square`
X ∼ B(4, 0.1)
`P(X=x)=""^n"C"_x p^x q^(n-x)= ""^4"C"_x (0.1)^x (0.9)^(4 - x)`
P[At most one defective device] = P[X ≤ 1]
= P[X=0] + P[X=1]
= `square+square`
∴ P[X ≤ 1] = `square`
A box contains 30 fruits, out of which 10 are rotten. Two fruits are selected at random one by one without replacement from the box. Find the probability distribution of the number of unspoiled fruits. Also find the mean of the probability distribution.
Five numbers x1, x2, x3, x4, x5 are randomly selected from the numbers 1, 2, 3, ......., 18 and are arranged in the increasing order such that x1 < x2 < x3 < x4 < x5. What is the probability that x2 = 7 and x4 = 11?
