Binomial Probability Problems

**Page 4**

31.

What is the probability that a leap year would have 52 Mondays and 53 Sundays?

a. | $\frac{1}{7}$ | ||

b. | $\frac{1}{52}$ | ||

c. | $\frac{2}{7}$ | ||

d. | $\frac{1}{53}$ |

P(E) =

A leap year has 366 days comprising 52 weeks & 2 days.

The above two days can be any one of the following 7 combinations. (i) Mon & Tue (ii) Tue & Wed (iii) Wed & Thurs (v) Thurs & Fri (v) Fri & Sat (vi) Sat & Sun (vii) Sun & Mon

Total number of outcomes n(S) = 7

If E were the Event of having 52 Mondays & 53 Sundays in the year, the only favorable case would be (vi) Sat & Sun

The total number of favorable cases n(E) = 1

P(E) =

[Substitute the values of n(E) and n(S) in P(E).]

Probability that a leap year would have 52 Mondays and 53 Sundays is

Correct answer : (1)

32.

A couple has 4 children. What is the probability that the children are 2 boys and 2 girls?

a. | $\frac{1}{5}$ | ||

b. | $\frac{3}{8}$ | ||

c. | $\frac{5}{16}$ | ||

d. | $\frac{3}{16}$ |

P(E) =

Let us denote B for boy and G for girl.

The sample space for the gender of children for a family that has four children is { BBBB, BBBG, BBGB, BBGG, BGBB, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG }.

Number of elements in sample space n(S) = 16

2 boys and 2 girls can be seen in { BBGG, BGBG, BGGB, GBBG, GBGB, GGBB }

Number of occurrences of 2 boys and 2 girls n(E) = 6

P(E) =

[Substitute the values of n(E) and n(S) in P(E).]

P(E) =

[Simplify.]

Probability that the children are 2 boys and 2 girls is

Correct answer : (2)

33.

Find the complement of the event: rolling a die and getting a 1 or 3.

a. | getting a 2, 4, 5 or 6 | ||

b. | getting a 5 or 3 | ||

c. | getting a 2 or 4 | ||

d. | getting a 2, 4 or 6 |

The sample space in rolling a die is {1, 2, 3, 4, 5, 6}.

Sample space included in the out come is {1, 3}.

Sample space not included in the out come is {2, 4, 5, 6}.

Complement of rolling a die and getting a 1 or 3 is getting a 2, 4, 5 or 6.

Correct answer : (1)

34.

How many elements are there in the complement of the event "selecting a month and getting a month that begins with J"?

a. | 1 | ||

b. | 9 | ||

c. | 7 | ||

d. | 3 |

[Number of months.]

Number of elements in the event of selecting a month starting with J = 3.

[January, June, July.]

Number of elements in the complement of the event of selecting a month and getting a month that begins with J = 12 - 3 = 9.

Correct answer : (2)

35.

Round the probabilities given:

I. 0.0003645.

II. 0.317.

III. 0.0193.

I. 0.0003645.

II. 0.317.

III. 0.0193.

a. | I 0.00036, II 0.32 and III 0.019. | ||

b. | I 0.0003, II 0.317 and III 0.019. | ||

c. | I 0.000365, II 0.317 and III 0.02. | ||

d. | I 0.0004, II 0.32 and III 0.02. |

So, 0.0003645 is rounded to 0.0004.

Probabilities can be rounded to 2 or 3 decimal places.

0.317 can be rounded to 0.32.

0.0193 can be rounded to 0.02.

So, the most appropriate rounding is, I 0.0004, II 0.32 and III 0.02.

Correct answer : (4)

36.

Select the correct statements:

I. An event consisting of a set of outcomes is a probability experiment.

II. A sample space is the set of all possible outcomes of a probability experiment.

III. An outcome is the result of a single trial of a probability experiment.

I. An event consisting of a set of outcomes is a probability experiment.

II. A sample space is the set of all possible outcomes of a probability experiment.

III. An outcome is the result of a single trial of a probability experiment.

a. | I and II only | ||

b. | I and III only | ||

c. | all are correct | ||

d. | II and III only |

Correct answer : (3)

37.

Select the correct statements:

I. P(E) + P($\stackrel{\u203e}{E}$) = 0.

II. If an event is certain, than the probability of that event is 1.

III. E $\cup $ $\stackrel{\u203e}{E}$ = S, where E is an event, $\stackrel{\u203e}{E}$ its complement and S the sample space.

IV. Number of events in the complement of selecting a king from a pack of cards is 50.

I. P(E) + P($\stackrel{\u203e}{E}$) = 0.

II. If an event is certain, than the probability of that event is 1.

III. E $\cup $ $\stackrel{\u203e}{E}$ = S, where E is an event, $\stackrel{\u203e}{E}$ its complement and S the sample space.

IV. Number of events in the complement of selecting a king from a pack of cards is 50.

a. | II and III only | ||

b. | I and III only | ||

c. | I and IV only | ||

d. | II, III and IV only |

Probability of an event which is certain to occur is 1.

E

Number of events in the complement of selecting a king from a pack of cards is 48.

Only statements II and III are correct.

Correct answer : (1)

38.

A box contains 5 red balls, 7 black balls and few white balls. If the probability of selecting a red ball from the box at a single draw is $\frac{1}{4}$, then how many white balls are there in the box?

a. | 20 | ||

b. | 12 | ||

c. | 8 | ||

d. | 7 |

Total number of balls in the box = 5 + 7 +

Probability of selecting a red ball =

Probability of selecting a red ball is given as

[Solve for

Number of white balls in the box = 8

Correct answer : (3)

39.

A box contains 5 red balls, 7 black balls and few white balls. If the probability of selecting a red ball from the box at a single draw is $\frac{1}{4}$, then what is the probability of selecting a white ball from the box in a single draw?

a. | $\frac{7}{20}$ | ||

b. | $\frac{2}{5}$ | ||

c. | $\frac{1}{4}$ | ||

d. | $\frac{3}{20}$ |

Total number of balls in the box = 5 + 7 +

Probability of selecting a red ball =

Probability of selecting a red ball is given as

[Solve for

Number of white balls in the box = 8

Total number of balls = 12 +

Probability of selecting a white ball =

Correct answer : (2)