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Question

Answers

*Extremely patient

*Extremely kind or honest.

Which of the above values do you prefer more?

Answer
Verified

Let us first discuss the formula of probability:

Probability is given by the number of favorable outcomes divided by total number of outcomes.

Now, let E be the favorable event whose probability we need to calculate and S be the sample space containing of all the possible outcomes, then

$P(E) = \dfrac{{n(E)}}{{n(S)}}$, where n(E) is the number of favorable outcomes and n(S) is the total number of possible outcomes.

Now let A be the set of extremely patient persons. Then, according to the question, we have: n(A) = 3.

Total number of persons is 12. So, n(S) = 12.

Let B be the set of extremely honest persons. Then, according to the question, we have: n(B) = 6.

Let C be the set of extremely kind persons. Then, according to the question, we have: $n(A) + n(B) + n(C) = n(S)$, so n(C) = 12 – 9 = 3.

Now, the first thing we need to do is find the probability of selecting a person who is extremely patient.

So, we need to find P(A).

Putting the values in the formula $P(E) = \dfrac{{n(E)}}{{n(S)}}$, where n(E) is the number of favorable outcomes and n(S) is the total number of possible outcomes. We will get:-

$P(A) = \dfrac{{n(A)}}{{n(S)}} = \dfrac{3}{{12}} = \dfrac{1}{4}$.

Now, we need to find the probability of selecting a person who is extremely kind or honest.

So, we need to find $P(B \cup C)$.

Putting the values in the formula $P(E) = \dfrac{{n(E)}}{{n(S)}}$, where n(E) is the number of favorable outcomes and n(S) is the total number of possible outcomes. We will get:-

$P(B \cup C) = \dfrac{{n(B \cup C)}}{{n(S)}} = \dfrac{{3 + 6}}{{12}} = \dfrac{9}{{12}} = \dfrac{3}{4}$.

Hence, we have the required values.

Now, according to me Honesty is the most important value because as long as a person is honest with himself/herself and others, he/she can find out where they are lacking and what they need to work on to grow better day by day. But other values are important as well.

Also notice that we took $B \cup C$ in the second part because we had “or” in the question. If it would have been an “and”, we would have taken intersection instead of union.