The variations occur due to the distinct rows of pennies
Response:
The probability that a student has a pet, given they do not have any siblings is:
Option: D ( 60%)
Step-by-step breakdown:
Let A represent the situation where a student lacks a sibling.
Let B signify the occurrence that a student has a pet.
Consequently, A∩B refers to the event in which a student is without siblings but possesses a pet.
Let P denote the chance of an event happening.
We need to determine:
P(B|A)
From our knowledge:

From the data provided:
P(A)=0.25
and P(A∩B)=0.15
Thus,

which expressed as a percentage is:

Therefore, the probability is:
60%
Assuming that the departure day for both is day 0, Sandy's next day off occurs on day 4, while Morgan's falls on day 10. Sandy's days off will be on days 4, 8, 12, 16, and 20. Meanwhile, Morgan's days off will be on days 10, 20, 30, 40, and 50. Thus, both of them will share a day off on day 20, meaning they will next have a joint day off in 20 days.
The likelihood that at least one trip occurs before Isabella's birth is 0.7627.
Step-by-step explanation:
In this scenario, Isabella has invented a time machine, but she lacks control over where she travels. Each use of the device holds a 0.25 probability of leading her to a time preceding her birth. Over the initial year of trials, she operates her machine 5 times. If we assume every journey has an equal chance of going back in time, we can calculate the odds that at least one of these trips occurs before she was born. Here's the calculation:
The probability of traveling to a time prior to her birth is 0.25.
The chance of not traveling back in time, given that the machine is used 5 times:
⇒ 
⇒ 
⇒ 
The probability that at least one trip goes before Isabella's birth is equal to 1 minus the probability of not traveling back to that period:
⇒ 
⇒ 
Consequently, the chance that at least one trip travels before Isabella's birth is 0.7627.