Question

Most armadillos of the genus Dasypus give birth to four monozygotic young (i.e., identical quadruplets).

Which statement is correct?

A random variable that could be studied while monitoring this species could be the number of male young born.

Because there are four babies, the probability of four males or four females must be 1/4 .

Because the young are all identical, there is no randomness.

In the long run, an equal number of males and females will be born.

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Parking is often a challenge, and it is also expensive to park on many college campuses. It is very tempting to park without paying, but there is a risk of getting a ticket. Based on empirical data, here is a probability table that quantifies the chances of various outcomes.

Event	No ticket	Warning	Fine	Towed
Probability	0.4	0.4	0.1	??

Which statement is not true?

Suppose you park for two days and that the probability above is the same for both days and each day’s outcome does not affect the other day’s outcome. The probability that you will have no ticket over this two‑day period is 0.36.

You parked on campus from Monday to Thursday and did not receive a ticket. Then the probability of no ticket on Friday (assuming each day’s events are independent) is still 0.40.

The probability of being towed is 0.10.

The probability that you will be fined or towed is 0.20.

Answer 1

Prob. 1:

A random variable that could be studied while monitoring this species could be the number of male young born: worng.

Because there are four babies, the probability of four males or four females must be 1/4 : Wrong

(Since armadillos of the genus Dasypus give birth to four monozygotic young, so each monozygotic young is either male or female. P(four males or four females)=P(four males)+P(four females)=1/16+1/16=1/8.)

Because the young are all identical, there is no randomness: Wrong (Young is either male or female).

Correct option: In the long run, an equal number of males and females will be born.

Prob. 2:

Since total probability=1 then P(No Ticket)+P(Warning)+P(Fine)+P(Towed)=1

or, 0.4+0.4+0.1+P(Towed)=1; or, P(Towed)=0.1

The probability that you will be fined or towed=P(fined)+P(Towed)= 0.20.

Option: Suppose you park for two days and that the probability above is the same for both days and each day’s outcome does not affect the other day’s outcome. The probability that you will have no ticket over this two‑day period is 0.36. (since correct answer=0.4*0.4=0.16)