Probability Distribution Worksheet

**Page 1**

1.

A fruit basket contains 5 plums, 4 apples and 3 oranges. Each time a fruit is picked at random, it is replaced in the basket. Find the probability that out of 6 fruits picked so, 3 are plums, 2 are apples and 1 is an orange.

a. | $\frac{25}{5184}$ | ||

b. | $\frac{125}{5184}$ | ||

c. | $\frac{25}{27}$ | ||

d. | $\frac{625}{5184}$ |

P(X =

where

Probability that the fruit is a plum P (Plum) =

P (Apple) =

P(Orange) =

P(3 plums, 2 apples, 1 orange) = [

[Substitute.]

=

[Simplify.]

So, the probability that out of 6 fruits picked so, 3 are plums, 2 are apples and 1 is an orange is

Correct answer : (4)

2.

A fair die is rolled 6 times. Find the probability that 1 appears exactly once, 2 appears exactly twice and 3 appears exactly thrice.

a. | $\frac{5}{3888}$ | ||

b. | $\frac{5}{324}$ | ||

c. | $\frac{5}{1944}$ | ||

d. | $\frac{5}{18}$ |

P(X =

where

Probability for any one number 1 or 2 or 3 to appear =

Setting

P(X = 1, X = 2, X = 3) = [

[Substitute and simplify.]

So, the probability that 1 appears exactly once, 2 appears exactly twice and 3 appears exactly thrice =

Correct answer : (1)

3.

In a cycle manufacturing factory, it is observed that chances that a cycle contains 0, 1, 2 defects is 0.88, 0.08, 0.04 respectively. If 16 cycles are chosen at random, then find the probability that 9 will have no defects, 4 have 1 defect and 3 have 2 defects.

a. | [$\frac{16!}{(9!4!3!)}$] × (0.88) ^{9} · (0.08)^{4} · (0.04)^{3} | ||

b. | [$\frac{16!}{(9!4!3!)}$] × (0.04) ^{3} · (0.88)^{9} · | ||

c. | (0.88) ^{9} · (0.08)^{4} · (0.04)^{3} | ||

d. | [$\frac{16!}{(9!4!3!)}$] × (0.88) ^{9} · (0.08)^{4} |

P(X =

where

Probability that there is no defect in the cycle = P (0 defect) = 0.88

Similarly, P (1 defect) = 0.08

P (2 defects) = 0.04

Setting

P (X = 0, X = 1, X = 2) = [

[Substitute and simplify.]

Correct answer : (1)

4.

The probability that a student commits 0, 1, 2, 3 errors in a short assignment is 0.6, 0.3, 0.05 and 0.05 respectively. If 10 assignments are chosen, then find the probability that 6 contain no errors, 2 contain 1 error, 1 contains 2 errors, and 1 contains 3 errors.

a. | $\frac{({10}^{3}\cdot {5}^{6})}{({3}^{10}\cdot 7)}$ | ||

b. | $\frac{({5}^{10}\cdot 7)}{({10}^{3}\cdot {3}^{6})}$ | ||

c. | $\frac{({3}^{10}\cdot 3)}{({5}^{3}\cdot {10}^{6})}$ | ||

d. | $\frac{({3}^{10}\cdot 7)}{({10}^{3}\cdot {5}^{6})}$ |

P(X =

where

Probability that there is no error in the assignment = P (0 error) = 0.6

Similarly, P (1 error) = 0.3

P (2 error) = 0.05

P (3 error) = 0.05

Setting

P(X) = [

[Substitute.]

P(X) =

[Simplify.]

So, the required probability is

Correct answer : (4)

5.

If 3% of the electric bulbs manufactured by a firm are defective, then find the probability that a sample of 100 bulbs has no defective bulb.

a. | $e$ ^{- 3} | ||

b. | 1 - $e$ ^{- 3} | ||

c. | 3$e$ ^{- 3} |

Number of samples

The situation is appropriate for making use of the Poisson distribution.

The Poisson variate mean, λ =

Substitute

P (X =

P (X = 0) =

[Substitute and simplify.]

So, a sample of 100 bulbs having no defective bulb has a probability of

Correct answer : (1)

6.

If 3% of the electric bulbs manufactured by a firm are defective, then what is the probability that exactly 5 bulbs out of a sample of 100 are defective?

a. | $\frac{81}{40{e}^{3}}$ | ||

b. | $\frac{27}{40{e}^{3}}$ | ||

c. | $\frac{81}{40{e}^{2}}$ | ||

d. | $\frac{9}{40{e}^{2}}$ |

Number of samples

The situation is appropriate for making use of the Poisson distribution.

The Poisson variate mean, λ =

Substitute

P (X =

P (X = 5) =

[Substitute and simplify.]

So, the probability that exactly 5 bulbs out of a sample of 100 are defective is

Correct answer : (1)

7.

Six coins are tossed 3200 times. Find the approximate probability of getting 5 heads 2 times.

a. | $\frac{{300}^{2}{e}^{-300}}{2}$ | ||

b. | $\frac{300{e}^{-300}}{200}$ | ||

c. | 300${e}^{-300}$ | ||

d. | $\frac{300{e}^{-300}}{2}$ |

Probability of getting a tail when a coin is tossed =

Probability of getting 5 heads with 6 coins

The situation is appropriate for applying the Poisson distribution with

The Poisson variate mean, λ =

Substitute

P (X =

P (X = 2) =

[Substitute and simplify.]

So, the probability of getting 5 heads 2 times is

Correct answer : (1)

8.

A manufacturing concern employing a large number of workers finds that over a period of time, the average absentee rate is 2 workers per shift of 100. Calculate the probability that exactly 2 workers will be absent in a shift.

a. | $\frac{1}{{e}^{2}}$ | ||

b. | $\frac{4}{{e}^{2}}$ | ||

c. | 2$e$ ^{2} | ||

d. | $\frac{2}{{e}^{2}}$ |

The situation is appropriate for use of Poisson distribution with

The Poisson variate mean, λ =

Substituting

P (X =

P (X = 2) =

[Substitute and simplify.]

So, the probability that exactly 2 workers will be absent in a shift is

Correct answer : (4)

9.

A company knows on the basis of its past experience that 2% of its blades are defective. Find the probability of having 3 defective blades in a sample of 100 blades. [Use $e$^{- 2} = 0.1353]

a. | 0.1804 | ||

b. | 0.2804 | ||

c. | 0.4804 | ||

d. | 0.3804 |

Applying the Poisson distribution,

The Poisson variate mean, λ =

Substituting

P (X =

P (X = 3) =

[Substitute and simplify.]

So, the probability of having 3 defective blades in a sample of 100 blades is 0.1804.

Correct answer : (1)

10.

The accidental drownings per year in a country is 3 per 100,000 population. Find the probability that in a city of 200,000 population, there will be fewer than 3 accidental drownings per year. [Use $e$^{- 6} = 0.002479]

a. | 0.0149 | ||

b. | 0.0025 | ||

c. | 0.062 | ||

d. | 0.0446 |

With

If there will be fewer than 3 accidental drownings per year, then substitute

P (X =

Required probability = P (X = 0) + P (X = 1) + P (X = 2)

Required probability =

[Substitute.]

Required probability = (1 + 6 + 18)(

[Simplify.]

So, the probability that there will be fewer than 3 accidental drownings per year is 0.062.

Correct answer : (3)