Probability and Conditional Probability MCQ Quiz - Objective Question with Answer for Probability and Conditional Probability - Download Free PDF

Explanation:

Given:

\(P(A) = P(\frac{A}{B}) = 0.25\)

and \(P(\frac{B}{A}) = 0.5\)

I. \(P(\frac{B}{A}) = \frac{P(A∩ B)}{P(A)}\)

⇒ P(A∩B) = P(A) P(B|A)

⇒ P(A∩B) = 0.25 × 0.5 = 0.125

Now

⇒ \(P(\frac{B}{A}) = \frac{P(A∩ B)}{P(B)}\)

⇒ \(P(B)= \frac{P(A∩ B)}{P(\frac{A}{B})}\)

⇒ \(P(B) = \frac{0.125}{0.25} = 0.5\)

Now, P(A).P(B) = 0.25 × 0.5 = 0.125 = P(A∩B)

Thus  A and B are independent

II. \(P(\overline A\cup \overline B ) = 1 – P(A ∩ B)\)

= 1 – 0.125 = 0.875

III. \(P(\overline A∩ \overline B ) = 1 – P(A \cup B)\)

= 1 – [P(A) + P(B) – P(A ∩ B)

=  1 – [0.25 + 0.5 – 0.125]

= = 1 – 0.625 = 0.375

So all statements I, II, and III are correct.

∴ Option (d) is correct.

The correct answer is Option 1.

Key Points

• Let the probability that A hits the target be P(A)=14" id="MathJax-Element-62-Frame" role="presentation" style="position: relative;" tabindex="0">P(A)=14P(A)=14 $P(A) = \frac{1}{4}$ .
• Let the probability that B hits the target be P(B)=25" id="MathJax-Element-63-Frame" role="presentation" style="position: relative;" tabindex="0">P(B)=25P(B)=25 $P(B) = \frac{2}{5}$ .
• The probability that neither A nor B hits the target is the product of their individual probabilities of missing the target.
• The probability that A misses the target is 1−P(A)=1−14=34" id="MathJax-Element-64-Frame" role="presentation" style="position: relative;" tabindex="0">1−P(A)=1−14=341−P(A)=1−14=34 $1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4}$ .
• The probability that B misses the target is 1−P(B)=1−25=35" id="MathJax-Element-65-Frame" role="presentation" style="position: relative;" tabindex="0">1−P(B)=1−25=351−P(B)=1−25=35 $1 - P(B) = 1 - \frac{2}{5} = \frac{3}{5}$ .
• The probability that neither hits the target is (34)×(35)=920" id="MathJax-Element-66-Frame" role="presentation" style="position: relative;" tabindex="0">(34)×(35)=920(34)×(35)=920 $\left(\frac{3}{4}\right) \times \left(\frac{3}{5}\right) = \frac{9}{20}$ .
• The probability that at least one of them hits the target is 1−920=1120" id="MathJax-Element-67-Frame" role="presentation" style="position: relative;" tabindex="0">1−920=11201−920=1120 $1 - \frac{9}{20} = \frac{11}{20}$ .

Additional Information

• This problem involves the concept of complementary probability, which is useful when calculating the probability of at least one event occurring.In probability theory, the complement rule is used to determine the probability of an event not occurring.
• Option 1 is indeed correct as the final probability calculation matches the probability required.

The probability of drawing a red card is 0.5. The probability of the complement event (not drawing a red card) is \(1 - 0.5 = 0.5\). Therefore, the probability of not drawing a red card is 0.5, which is option 1.

To find the probability of drawing a red or green ball, we calculate the total number of red and green balls and divide it by the total number of balls. There are 10 red balls and 6 green balls, totaling 16. The total number of balls is 10 + 14 + 6 = 30. Therefore, the probability is \( \frac{16}{30} = \frac{8}{15} \). Option 1 only considers blue balls, Option 2 only considers green balls, and Option 4 only considers red balls.

The even hours between 1 and 14 are 2, 4, 6, 8, 10, 12, and 14, giving us 7 even numbers. The total number of hours is 14. Hence, the probability of the clock playing a tune corresponding to an even hour is \( \frac{7}{14} \), which simplifies to \( \frac{1}{2} \). The options 5/14, 6/14, and 8/14 do not correctly represent the count of even numbers.

Explanation:

Given:

\(P(A) = P(\frac{A}{B}) = 0.25\)

and \(P(\frac{B}{A}) = 0.5\)

I. \(P(\frac{B}{A}) = \frac{P(A∩ B)}{P(A)}\)

⇒ P(A∩B) = P(A) P(B|A)

⇒ P(A∩B) = 0.25 × 0.5 = 0.125

Now

⇒ \(P(\frac{B}{A}) = \frac{P(A∩ B)}{P(B)}\)

⇒ \(P(B)= \frac{P(A∩ B)}{P(\frac{A}{B})}\)

⇒ \(P(B) = \frac{0.125}{0.25} = 0.5\)

Now, P(A).P(B) = 0.25 × 0.5 = 0.125 = P(A∩B)

Thus  A and B are independent

II. \(P(\overline A\cup \overline B ) = 1 – P(A ∩ B)\)

= 1 – 0.125 = 0.875

III. \(P(\overline A∩ \overline B ) = 1 – P(A \cup B)\)

= 1 – [P(A) + P(B) – P(A ∩ B)

=  1 – [0.25 + 0.5 – 0.125]

= = 1 – 0.625 = 0.375

So all statements I, II, and III are correct.

∴ Option (d) is correct.

The correct answer is Option 1.

Key Points

• Let the probability that A hits the target be P(A)=14" id="MathJax-Element-62-Frame" role="presentation" style="position: relative;" tabindex="0">P(A)=14P(A)=14 $P(A) = \frac{1}{4}$ .
• Let the probability that B hits the target be P(B)=25" id="MathJax-Element-63-Frame" role="presentation" style="position: relative;" tabindex="0">P(B)=25P(B)=25 $P(B) = \frac{2}{5}$ .
• The probability that neither A nor B hits the target is the product of their individual probabilities of missing the target.
• The probability that A misses the target is 1−P(A)=1−14=34" id="MathJax-Element-64-Frame" role="presentation" style="position: relative;" tabindex="0">1−P(A)=1−14=341−P(A)=1−14=34 $1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4}$ .
• The probability that B misses the target is 1−P(B)=1−25=35" id="MathJax-Element-65-Frame" role="presentation" style="position: relative;" tabindex="0">1−P(B)=1−25=351−P(B)=1−25=35 $1 - P(B) = 1 - \frac{2}{5} = \frac{3}{5}$ .
• The probability that neither hits the target is (34)×(35)=920" id="MathJax-Element-66-Frame" role="presentation" style="position: relative;" tabindex="0">(34)×(35)=920(34)×(35)=920 $\left(\frac{3}{4}\right) \times \left(\frac{3}{5}\right) = \frac{9}{20}$ .
• The probability that at least one of them hits the target is 1−920=1120" id="MathJax-Element-67-Frame" role="presentation" style="position: relative;" tabindex="0">1−920=11201−920=1120 $1 - \frac{9}{20} = \frac{11}{20}$ .

Additional Information

• This problem involves the concept of complementary probability, which is useful when calculating the probability of at least one event occurring.In probability theory, the complement rule is used to determine the probability of an event not occurring.
• Option 1 is indeed correct as the final probability calculation matches the probability required.

Choice B is correct. According to the table, a total of 671 customers asked about a bill. Of these, 48 also asked for repairs. Therefore, if a customer who asked about a bill is selected at random, the probability that the customer also asked for repairs is \(\frac{48}{671} \approx 0.07\). 
Choice A is incorrect. This is the probability that a customer selected at random from all customers who called on Monday both asked for repairs and asked about a bill. Choice C is incorrect. This is the probability that a customer selected at random from all customers who called on Monday asked for repairs, regardless of whether or not the customer asked about a bill. Choice D is incorrect. This is the probability that a customer selected at random from those who asked for repairs also asked about a bill. 

The correct answer is \(\frac{23}{60}\). It's given that one of the participants will be selected at random. The probability of selecting a participant from group A given that the participant is at least 10 years of age is the number of participants in group A who are at least 10 years of age divided by the total number of participants who are at least 10 years of age. The table shows that in group A, there are 14 participants who are 10 - 19 years of age and 9 participants who are 20+ years of age. Therefore, there are 14 + 9, or 23, participants in group A who are at least 10 years of age. The table also shows that there are a total of 30 participants who are 10 - 19 years of age and 30 participants who are 20+ years of age. Therefore, there are a total of 30 + 30, or 60, participants who are at least 10 years of age. It follows that the probability of selecting a participant from group A given that the participant is at least 10 years of age is \(\frac{23}{60}\). Note that 23/60, .3833, and 0.383 are examples of ways to enter a correct answer.

The correct answer is 42. It’s given that in Nigeria on September 30, 2015, the probability of selecting an MTN subscriber from all Internet subscribers is 0.43, that there were p million, or p(1,000,000), MTN subscribers, and that there were 97 million, or 97,000,000, Internet subscribers. The probability of selecting an MTN subscriber from all Internet subscribers can be found by dividing the number of MTN subscribers by the total number of Internet subscribers. Therefore, the equation \(\frac{p(1,000,000)}{97,000,000}=0.43\) can be used to solve for p. Dividing 1,000,000 from the numerator and denominator of the expression on the left-hand side yields \(\frac{p}{97}=0.43\). Multiplying both sides of this equation by 97 yields p = (0.43)(97) = 41.71, which, to the nearest integer, is 42.

The correct answer is \(\frac{1}{3}\). It’s given that the number of email responses is twice the number of phone responses. Therefore, if the number of phone responses is p, then the number of email responses is 2p. The table shows that 20% of people who responded by phone preferred a picnic. It follows that the expression 0.20p represents the number of these people. The table also shows that 5% of the people who responded by email preferred a picnic. The expression 0.05(2p), or 0.1p, represents the number of these people. Therefore, a total of 0.20p + 0.1p, or 0.3p people preferred a picnic. Thus, the probability of selecting at random a person who responded by email from the people who preferred a picnic is \(\frac{0.1 p}{0.3 p} \text {, or } \frac{1}{3}\). Note that 1/3, .3333, and 0.333 are examples of ways to enter a correct answer. 

The correct answer is 8. In this group, \(\frac{1}{9}\) of the people who are rhesus negative have blood type B. The total number of people who are rhesus negative in the group is 7 + 2 + 1 + x, and there are 2 people who are rhesus negative with blood type B. Therefore, \(\frac{2}{(7+2+1+x)}=\frac{1}{9}\). Combining like terms on the left-hand side of the equation yields \(\frac{2}{(10+x)}=\frac{1}{9}\). Multiplying both sides of this equation by 9 yields \(\frac{18}{(10+x)}=1\), and multiplying both sides of this equation by (10 + x) yields 18 = 10 + x. Subtracting 10 from both sides of this equation yields 8 = x. 

The correct answer is \(\frac{5}{7}\). It is given that no contestant received the same score on two different days, so each of the contestants who received a score of 5 is represented in the “5 out of 5” column of the table exactly once. Therefore, the probability of selecting a contestant who received a score of 5 on Day 2 or Day 3, given that the contestant received a score of 5 on one of the three days, is found by dividing the total number of contestants who received a score of 5 on Day 2 or Day 3 (2 + 3 = 5) by the total number of contestants who received a score of 5, which is given in the table as 7. So the probability is \(\frac{5}{7}\). Note that 5/7, .7142, .7143, and 0.714 are examples of ways to enter a correct answer. 

Choice B is correct. In total, there are 6 people in the Human Resources department. Of those 6, 2 have a master’s degree as their highest level of education. Therefore, the probability of an employee selected at random from the Human Resources department having a master’s degree is \(\frac{2}{6}\), which simplifies to \(\frac{1}{3}\). 
Choice A is incorrect; it is the probability that an employee selected at random from either department will be in the Human Resources department and have a master’s degree. Choice C is incorrect; it is the probability that an employee with a master’s degree selected at random will be in the Human Resources department. Choice D is incorrect; it is the probability that an employee selected at random from either department will have a master’s degree.

The correct answer is \(\frac{13}{50}\). It's given that a box contains 13 red pens and 37 blue pens. If one of these pens is selected at random, the probability of selecting a red pen is the number of red pens in the box divided by the number of red and blue pens in the box. The number of red and blue pens in the box is 13 + 37, or 50. Since there are 13 red pens in the box, it follows that the probability of selecting a red pen is \(\frac{13}{50}\). Note that 13/50 and .26 are examples of ways to enter a correct answer.