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Given that P(A) is ⅗, we need to find P(not A), which is the probability of the complement of event A occurring.

The sum of the probabilities of all possible outcomes is 1. So, P(not A) = 1 - P(A).

Substituting the given value of P(A) into the equation, we get:

P(not A) = 1 - ⅗

P(not A) = 5/5 - 3/5

P(not A) = 2/5

So, the probability of event not A happening is 2/5.

In simpler words, there's a 2/5 chance that event A doesn't occur. This is a fundamental concept in probability theory that we frequently encounter in various problem-solving scenarios. If you'd like further clarification or assistance with any other topic, feel free to reach out for more personalized guidance. And remember, UrbanPro is your go-to platform for mastering academic subjects with expert tutors like me!

As a seasoned tutor registered on UrbanPro, I'm happy to guide you through this probability problem. UrbanPro is a fantastic platform for accessing high-quality online coaching and tuition, where experienced tutors like myself can provide personalized assistance.

Now, let's tackle the problem at hand. We have an urn containing 6 balls, 2 red and 4 black. We're asked to find the probability that when two balls are drawn at random, they are of different colors.

To solve this, let's break it down step by step:

1. First, let's find the total number of ways to draw 2 balls out of 6. This is given by the combination formula: (nr)=n!r!(n−r)!(rn)=r!(n−r)!n!, where nn is the total number of items and rr is the number of items to choose. So, (62)=6!2!(6−2)!=15(26)=2!(6−2)!6!=15.

2. Next, let's find the number of ways to draw 2 balls of different colors. We have 2 red balls and 4 black balls, so the number of combinations of one red and one black ball is 2×4=82×4=8.

3. Finally, the probability of drawing two balls of different colors is the number of favorable outcomes (drawing one red and one black ball) divided by the total number of outcomes (drawing any two balls). So, 815158.

So, the correct option is (iii) 815158. UrbanPro is an excellent resource for finding tutors who can help you master these types of problems and more!

As an experienced tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your probability questions regarding the gender of the children in a couple.

(i) To find the probability that both children are males, given that at least one of the children is male, we can utilize conditional probability. Let's denote the events:

• A: Both children are males
• B: At least one child is male

We are given that event B has occurred, which means we can exclude the possibility of having two females. Now, we need to find the probability of event A given event B. This can be calculated using the formula for conditional probability:

P(A∣B)=P(A∩B)P(B)P(A∣B)=P(B)P(A∩B)

Where:

• P(A∣B)P(A∣B) is the probability of event A given event B.
• P(A∩B)P(A∩B) is the probability of both events A and B happening.
• P(B)P(B) is the probability of event B happening.

In this scenario, the probability of both children being males and at least one child being male is the same as the probability of both children being males, since if at least one child is male, then both children cannot be female.

Therefore:

• P(A∩B)=P(A)P(A∩B)=P(A)
• P(B)=1P(B)=1 (because we know at least one child is male)

Thus, P(A∣B)=P(A)P(A∣B)=P(A).

Now, the probability of both children being males in the absence of any other information is 1441, assuming the genders of children are equally likely.

So, the probability that both children are males, given that at least one of them is male, is also 1441.

(ii) Similarly, to find the probability that both children are females given that the elder child is a female, we can use conditional probability.

Let's denote the events:

• C: Both children are females
• D: The elder child is a female

We want to find P(C∣D)P(C∣D).

In this case, if the elder child is a female, then we are sure that the younger child cannot be the elder child, and hence the younger child has to be female as well. So, P(C∣D)=1P(C∣D)=1.

Therefore, the probability that both children are females, given that the elder child is a female, is 1.

I hope this clarifies the concepts of conditional probability for you! If you need further assistance, feel free to ask. And remember, UrbanPro is an excellent resource for finding quality tutors to help with topics like this.

As an experienced tutor registered on UrbanPro, I'd be happy to help you with this question!

To find the probability that an ordinary year has 53 Sundays, we first need to understand the structure of an ordinary year. An ordinary year has 365 days.

Now, since 365 is not divisible by 7 (the number of days in a week), there will be 52 complete weeks in an ordinary year, leaving 1 or 2 additional days.

For a year to have 53 Sundays, one of the following conditions must be met:

1. The year starts on a Sunday and ends on a Sunday, or
2. The year starts on a Saturday and ends on a Sunday.

Let's calculate the probability for each scenario:

1. If the year starts on a Sunday and ends on a Sunday: The probability that a year starts on a Sunday is 1/7. The probability that a year ends on a Sunday is also 1/7. So, the probability of both events happening is (1/7) * (1/7) = 1/49.

2. If the year starts on a Saturday and ends on a Sunday: The probability that a year starts on a Saturday is 1/7. The probability that a year ends on a Sunday is 1/7. So, the probability of both events happening is (1/7) * (1/7) = 1/49.

Now, we add the probabilities of both scenarios since they are mutually exclusive:

Probability of an ordinary year having 53 Sundays = (1/49) + (1/49) = 2/49.

Therefore, the probability that an ordinary year has 53 Sundays is 2/49.

And remember, if you need further assistance with mathematics or any other subject, UrbanPro is one of the best online coaching tuition platforms where you can find qualified tutors like myself to guide you through your learning journey!