Binomial Probability Problems - Page 4

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. $\frac{1}{53}$

Solution:

Probability of event E : P(E) = Number of outcomes in E / Total number of outcomes in the sample space
P(E) = n(E)n(S)

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) = 1 / 7
[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 1 / 7.

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. $\frac{3}{16}$

Solution:

Probability of event E : P(E) = Number of outcomes in E / Total number of outcomes in the sample space
P(E) = n(E)n(S)

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) = 6 / 16
[Substitute the values of n(E) and n(S) in P(E).]

P(E) = 3 / 8
[Simplify.]

Probability that the children are 2 boys and 2 girls is 3 / 8.

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

Solution:

Complement of an event E is the set of out comes in the sample space that are not included in the outcomes of that event E.

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.

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

Solution:

Number of elements in the sample space event of selecting a month = 12.
[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.

35.
Round the probabilities given:
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.

Solution:

When probability of an event is an extremely small decimal, the decimal is rounded to the first non zero digit.

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.

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.
 a. I and II only b. I and III only c. all are correct d. II and III only

Solution:

All the statements are correct.

37.
Select the correct statements:
I. P(E) + P($\stackrel{‾}{E}$) = 0.
II. If an event is certain, than the probability of that event is 1.
III. E $\cup$ $\stackrel{‾}{E}$ = S, where E is an event, $\stackrel{‾}{E}$ 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

Solution:

P(E) + P(E) = 1.

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

E E = S.

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

Only statements II and III are correct.

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

Solution:

Let the number of white balls be x

Total number of balls in the box = 5 + 7 + x = 12 + x

Probability of selecting a red ball = number of red ballstotal number of balls = 512+x

Probability of selecting a red ball is given as 1 / 4.

512+x = 1 / 4

x = 8
[Solve for x.]

Number of white balls in the box = 8

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. $\frac{3}{20}$

Solution:

Let the number of white balls be x

Total number of balls in the box = 5 + 7 + x = 12 + x

Probability of selecting a red ball = number of red ballstotal number of balls = 512+x

Probability of selecting a red ball is given as 1 / 4.

512+x = 1 / 4

x = 8
[Solve for x.]

Number of white balls in the box = 8

Total number of balls = 12 + x = 20

Probability of selecting a white ball = 8 / 20 = 2 / 5