###### Advertisements

###### Advertisements

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as number greater than 4 appears on at least one die.

###### Advertisements

#### Solution 1

When a die is tossed twice, the sample space S has 6 × 6 = 36 sample points.

∴ n(S) = 36

Trial will be a success if the number on at least one die is 5 or 6.

Let X denote the number of dice on which 5 or 6 appears.

Then X can take values 0, 1, 2

When X = 0 i.e., 5 or 6 do not appear on any of the dice, then

X = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}.

∴ n(X) = 16

∴ P(X = 0) =`(n(X))/(n(S)) = 16/36 = 4/9`

When X = 1, i.e. 5 or 6 appear on exactly one of the dice, then

X = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}

∴ n(X) = 16

∴ P(X = 1) = `(n(X))/(n(S)) = 16/36 = 4/9`

When X = 2, i.e. 5 or 6 appear on both of the dice, then

X = {(5, 5), (5, 6), (6, 5), (6, 6)}

∴ n(X) = 4

∴ P(X = 2) =`(n(X))/(n(S)) = 4/36 = 1/9`

∴ The required probability distribution is

X |
0 | 1 | 2 |

P(X = x) |
`4/9` | `4/9` | `1/9` |

#### Solution 2

Success is defined as a number greater than 4 appears on at least one die

Let X denote the number of successes.

∴ Possible values of X and 0, 1, 2.

Let P(getting a number greater than 4) = p

= `2/6`

= `1/3`

∴ q = 1 – p

= `1 - 1/3`

= `2/3`

∴ P(X = 0) = P(no success)

= q^{2}

= `4/9`

P(X = 1) = P(one success)

= qp + pq = 2pq

= `2 xx 1/3 xx 2/3`

= `4/9`

P(X = 2) = P(two successes)

= pp

= p^{2}

= `1/9`

∴ Probability distribution of X is as follows:

X |
0 | 1 | 2 |

P(X = x) |
`4/9` | `4/9` | `1/9` |

#### APPEARS IN

#### RELATED QUESTIONS

State if the following is not the probability mass function of a random variable. Give reasons for your answer.

X |
0 | 1 | 2 |

P(X) |
0.1 | 0.6 | 0.3 |

State if the following is not the probability mass function of a random variable. Give reasons for your answer.

Y |
−1 | 0 | 1 |

P(Y) |
0.6 | 0.1 | 0.2 |

State if the following is not the probability mass function of a random variable. Give reasons for your answer.

X | 0 | -1 | -2 |

P(X) | 0.3 | 0.4 | 0.3 |

**A random variable X has the following probability distribution:**

X |
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

P(X) |
0 | k | 2k | 2k | 3k | k^{2} |
2k^{2} |
7k^{2} + k |

**Determine:**

- k
- P(X < 3)
- P( X > 4)

Find expected value and variance of X for the following p.m.f.

x |
-2 | -1 | 0 | 1 | 2 |

P(X) |
0.2 | 0.3 | 0.1 | 0.15 | 0.25 |

The following is the p.d.f. of r.v. X:

f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise.

Find P (x < 1·5)

The following is the p.d.f. of r.v. X:

f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise.

P(x > 2)

It is known that error in measurement of reaction temperature (in 0° c) in a certain experiment is continuous r.v. given by

f (x) = `x^2 /3` , for –1 < x < 2 and = 0 otherwise

Verify whether f (x) is p.d.f. of r.v. X.

It is known that error in measurement of reaction temperature (in 0° c) in a certain experiment is continuous r.v. given by

f (x) = `x^2/3` , for –1 < x < 2 and = 0 otherwise

Find probability that X is negative

Find k if the following function represent p.d.f. of r.v. X

f (x) = kx, for 0 < x < 2 and = 0 otherwise, Also find P `(1/ 4 < x < 3 /2)`.

Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by

f (x) = `1/ 5` , for 0 ≤ x ≤ 5 and = 0 otherwise.

Find the probability that waiting time is between 1 and 3

If a r.v. X has p.d.f.,

f (x) = `c /x` , for 1 < x < 3, c > 0, Find c, E(X) and Var (X).

**Choose the correct option from the given alternative:**

If a d.r.v. X takes values 0, 1, 2, 3, . . . which probability P (X = x) = k (x + 1)·5 ^{−x} , where k is a constant, then P (X = 0) =

**Choose the correct option from the given alternative:**

If p.m.f. of a d.r.v. X is P (X = x) = `((c_(x)^5 ))/2^5` , for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise If a = P (X ≤ 2) and b = P (X ≥ 3), then E (X ) =

**Choose the correct option from the given alternative:**

If p.m.f. of a d.r.v. X is P (X = x) = `x^2 /(n (n + 1))`, for x = 1, 2, 3, . . ., n and = 0, otherwise then E (X ) =

**Choose the correct option from the given alternative :**

If p.m.f. of a d.r.v. X is P (x) = `c/ x^3` , for x = 1, 2, 3 and = 0, otherwise (elsewhere) then E (X ) =

**Solve the following :**

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

Amount of syrup prescribed by physician.

**Solve the following problem :**

A fair coin is tossed 4 times. Let X denote the number of heads obtained. Identify the probability distribution of X and state the formula for p. m. f. of X.

The following is the c.d.f. of r.v. X

x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |

F(X) | 0.1 | 0.3 | 0.5 | 0.65 | 0.75 | 0.85 | 0.9 |
1 |

P (X ≤ 3/ X > 0)

Let X be amount of time for which a book is taken out of library by randomly selected student and suppose X has p.d.f

f (x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise.

Calculate: P(x≤1)

Let X be amount of time for which a book is taken out of library by randomly selected student and suppose X has p.d.f

f (x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise.

Calculate: P(0.5 ≤ x ≤ 1.5)

Find the probability distribution of number of number of tails in three tosses of a coin

Find the probability distribution of number of heads in four tosses of a coin

70% of the members favour and 30% oppose a proposal in a meeting. The random variable X takes the value 0 if a member opposes the proposal and the value 1 if a member is in favour. Find E(X) and Var(X).

Find k if the following function represents the p. d. f. of a r. v. X.

f(x) = `{(kx, "for" 0 < x < 2),(0, "otherwise."):}`

Also find `"P"[1/4 < "X" < 1/2]`

Given that X ~ B(n, p), if n = 10 and p = 0.4, find E(X) and Var(X)

Given that X ~ B(n,p), if n = 10, E(X) = 8, find Var(X).

X is r.v. with p.d.f. f(x) = `"k"/sqrt(x)`, 0 < x < 4 = 0 otherwise then x E(X) = _______

**Solve the following problem :**

Find the expected value and variance of the r. v. X if its probability distribution is as follows.

x |
– 1 | 0 | 1 |

P(X = x) |
`(1)/(5)` | `(2)/(5)` | `(2)/(5)` |

**Solve the following problem :**

Let the p. m. f. of the r. v. X be

`"P"(x) = {((3 - x)/(10)", ","for" x = -1", "0", "1", "2.),(0,"otherwise".):}`

Calculate E(X) and Var(X).

**Solve the following problem :**

Let X∼B(n,p) If E(X) = 5 and Var(X) = 2.5, find n and p.

If X denotes the number on the uppermost face of cubic die when it is tossed, then E(X) is ______

If the p.m.f. of a d.r.v. X is P(X = x) = `{{:(x/("n"("n" + 1))",", "for" x = 1"," 2"," 3"," .... "," "n"),(0",", "otherwise"):}`, then E(X) = ______

If the p.m.f. of a d.r.v. X is P(X = x) = `{{:(("c")/x^3",", "for" x = 1"," 2"," 3","),(0",", "otherwise"):}` then E(X) = ______

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

X |
–2 | –1 | 0 | 1 | 2 | 3 |

P(X = x) |
0.1 | k | 0.2 | 2k | 0.3 | k |

then P(X = –1) is ______

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

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

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

then k = ______

Find mean for the following probability distribution.

X |
0 | 1 | 2 | 3 |

P(X = x) |
`1/6` | `1/3` | `1/3` | `1/6` |

Find the expected value and variance of r.v. X whose p.m.f. is given below.

X |
1 | 2 | 3 |

P(X = x) |
`1/5` | `2/5` | `2/5` |

**The probability distribution of X is as follows:**

X |
0 | 1 | 2 | 3 | 4 |

P(X = x) |
0.1 | k | 2k | 2k | k |

Find k and P[X < 2]

**Choose the correct alternative:**

f(x) is c.d.f. of discete r.v. X whose distribution is

x_{i} |
– 2 | – 1 | 0 | 1 | 2 |

p_{i} |
0.2 | 0.3 | 0.15 | 0.25 | 0.1 |

then F(– 3) = ______

If p.m.f. of r.v. X is given below.

x |
0 | 1 | 2 |

P(x) |
q^{2} |
2pq | p^{2} |

then Var(x) = ______

If X is discrete random variable takes the values x_{1}, x_{2}, x_{3}, … x_{n}, then `sum_("i" = 1)^"n" "P"(x_"i")` = ______

E(x) is considered to be ______ of the probability distribution of x.

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

x |
1 | 2 | 3 | 4 | 5 | 6 |

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

Complete the following activity.

**Solution:** Since `sum"p"_"i"` = 1

k = `square`

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

x |
1 | 2 | 3 | 4 | 5 | 6 |

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

Complete the following activity.

**Solution:** Since `sum"p"_"i"` = 1

P(X ≤ 4) = `square + square + square + square = square`

Using the following activity, find the expected value and variance of the r.v.X if its probability distribution is as follows.

x |
1 | 2 | 3 |

P(X = x) |
`1/5` | `2/5` | `2/5` |

**Solution:** µ = E(X) = `sum_("i" = 1)^3 x_"i""p"_"i"`

E(X) = `square + square + square = square`

Var(X) = `"E"("X"^2) - {"E"("X")}^2`

= `sum"X"_"i"^2"P"_"i" - [sum"X"_"i""P"_"i"]^2`

= `square - square`

= `square`

The following function represents the p.d.f of a.r.v. X

f(x) = `{{:((kx;, "for" 0 < x < 2, "then the value of K is ")),((0;, "otherwise")):}` ______

**The probability distribution of a discrete r.v. X is as follows:**

x |
1 | 2 | 3 | 4 | 5 | 6 |

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

- Determine the value of k.
- Find P(X ≤ 4)
- P(2 < X < 4)
- P(X ≥ 3)

**The p.m.f. of a random variable X is as follows:**

P (X = 0) = 5k^{2}, P(X = 1) = 1 – 4k, P(X = 2) = 1 – 2k and P(X = x) = 0 for any other value of X. Find k.

Given below is the probability distribution of a discrete random variable x.

X |
1 | 2 | 3 | 4 | 5 | 6 |

P(X = x) |
K | 0 | 2K | 5K | K | 3K |

Find K and hence find P(2 ≤ x ≤ 3)