Advertisements
Advertisements
प्रश्न
Two probability distributions of the discrete random variable X and Y are given below.
| X | 0 | 1 | 2 | 3 |
| P(X) | `1/5` | `2/5` | `1/5` | `1/5` |
| Y | 0 | 1 | 2 | 3 |
| P(Y) | `1/5` | `3/10` | `2/10` | `1/10` |
Prove that E(Y2) = 2E(X).
उत्तर
First probability distribution is given by
| X | 0 | 1 | 2 | 3 |
| P(X) | `1/5` | `2/5` | `1/5` | `1/5` |
We know that, E(X) = `sum_("i" = 11)^"n" "P"_"i""X"_"i"`
⇒ E(X) = `0 * 1/5 + 1 * 2/5 + 2 * 1/5 + 3 * 1/5`
= `0 + 2/5 + 2/5 + 3/5`
= `7/5`
For the second probability distribution,
| Y | 0 | 1 | 2 | 3 |
| P(Y) | `1/5` | `3/10` | `2/10` | `1/10` |
E(Y2) = `0 * 1/5 + 1 * 3/10 + 4 * 2/5 + 9 * 1/10`
= `0 + 3/10 + 8/5 + 9/10`
= `28/10`
= `14/5`
Now E(Y2) = `14/5` and 2E(X) = `2*7/5 = 14/5`
Hence, E(Y2) = 2E(X).
APPEARS IN
संबंधित प्रश्न
A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
The random variable X has probability distribution P(X) of the following form, where k is some number:
`P(X = x) {(k, if x = 0),(2k, if x = 1),(3k, if x = 2),(0, "otherwise"):}`
- Determine the value of 'k'.
- Find P(X < 2), P(X ≥ 2), P(X ≤ 2).
The probability distribution function of a random variable X is given by
| xi : | 0 | 1 | 2 |
| pi : | 3c3 | 4c − 10c2 | 5c-1 |
where c > 0 Find: c
Three cards are drawn successively with replacement from a well-shuffled deck of 52 cards. A random variable X denotes the number of hearts in the three cards drawn. Determine 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?
A discrete random variable X has the probability distribution given below:
| X: | 0.5 | 1 | 1.5 | 2 |
| P(X): | k | k2 | 2k2 | k |
Find the value of k.
A discrete random variable X has the probability distribution given below:
| X: | 0.5 | 1 | 1.5 | 2 |
| P(X): | k | k2 | 2k2 | k |
Determine the mean of the distribution.
A random variable X takes the values 0, 1, 2, 3 and its mean is 1.3. If P (X = 3) = 2 P (X = 1) and P (X = 2) = 0.3, then P (X = 0) 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
Mark the correct alternative in the following question:
For the following probability distribution:
| X: | −4 | −3 | −2 | −1 | 0 |
| P(X): | 0.1 | 0.2 | 0.3 | 0.2 | 0.2 |
The value of E(X) is
A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of number of successes.
Compute the age specific death rate for the following data :
| Age group (years) | Population (in thousands) | Number of deaths |
| Below 5 | 15 | 360 |
| 5-30 | 20 | 400 |
| Above 30 | 10 | 280 |
Verify whether the following function can be regarded as probability mass function (p.m.f.) for the given values of X :
| X | -1 | 0 | 1 |
| P(X = x) | -0.2 | 1 | 0.2 |
Determine whether each of the following is a probability distribution. Give reasons for your answer.
| x | 0 | 1 | 2 | 3 | 4 |
| P(x) | 0.1 | 0.5 | 0.2 | –0.1 | 0.3 |
Determine whether each of the following is a probability distribution. Give reasons for your answer.
| x | 0 | 1 | 2 |
| P(x) | 0.3 | 0.4 | 0.2 |
A sample of 4 bulbs is drawn at random with replacement from a lot of 30 bulbs which includes 6 defective bulbs. Find the probability distribution of the number of defective bulbs.
A coin is biased so that the head is 3 times as likely to occur as tail. Find the probability distribution of number of tails in two tosses.
A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of 2 successes
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
Solve the following problem :
If a fair coin is tossed 4 times, find the probability that it shows head in the first 2 tosses and tail in last 2 tosses.
Solve the following problem :
The probability that a component will survive a check test is 0.6. Find the probability that exactly 2 of the next 4 components tested survive.
Find the probability distribution of the number of doublets in three throws of a pair of dice
Let X be a discrete random variable. The probability distribution of X is given below:
| X | 30 | 10 | – 10 |
| P(X) | `1/5` | `3/10` | `1/2` |
Then E(X) is equal to ______.
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.
For the following probability distribution:
| X | 1 | 2 | 3 | 4 |
| P(X) | `1/10` | `3/10` | `3/10` | `2/5` |
E(X2) is equal to ______.
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
If the p.m.f of a r. v. X is
P(x) = `c/x^3`, for x = 1, 2, 3
= 0, otherwise
then E(X) = ______.
Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a nonprime number. The probability that the card was drawn from Box I is ______.
The probability that a bomb will hit the target is 0.8. Complete the following activity to find, the probability that, out of 5 bombs exactly 2 will miss the target.
Solution: Here, n = 5, X =number of bombs that hit the target
p = probability that bomb will hit the target = `square`
∴ q = 1 - p = `square`
Here, `X∼B(5,4/5)`
∴ P(X = x) = `""^"n""C"_x"P"^x"q"^("n" - x) = square`
P[Exactly 2 bombs will miss the target] = P[Exactly 3 bombs will hit the target]
= P(X = 3)
=`""^5"C"_3(4/5)^3(1/5)^2=10(4/5)^3(1/5)^2`
∴ P(X = 3) = `square`
