Question

For a one-week period, three bus routes were observed. The results are shown in the table below.

On-Time Delayed Total
Bus Route A 28 7 35
Bus Route B 37 8 45
Bus Route C 24 6 30
Total 89 21 110
A bus is selected randomly. Which event has the highest probability?

A.
The bus is from route B and is delayed.
B.
The bus is from route A and is on time.
C.
The bus is from route C and is on time.
D.
The bus is from route C and is delayed.

Answer 1

Using it's concept, the event that has the highest probability is given as follows:

B. The bus is from route A and is on time.

What is a probability?

A probability is given by the number of desired outcomes divided by the number of total outcomes.

We have that out of the 110 buses, 8 are delayed from route B, hence:

P(A) = 8/110.

28 are on time from route A, hence:

P(B) = 28/110.

24 are on time from route C, hence:

P(C) = 24/110.

6 are delayed from route C, hence:
P(D) = 6/110.

Hence option B gives the highest probability.

More can be learned about probabilities at https://brainly.com/question/14398287

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Answer 2

Answer:

B.  The bus is from route A and is on time.

Step-by-step explanation:

$\large \begin{array}{| c | c | c | c |}\cline{1-4} & \sf On-Time & \sf Delayed & \sf Total\\\cline{1-4} \sf Bus\:Route\:A & 28 & 7 & 35\\\cline{1-4} \sf Bus\:Route\:B & 37 & 8 & 45\\\cline{1-4} \sf Bus\:Route\:C & 24 & 6 & 30\\\cline{1-4} \sf Total & 89 & 21 & 110\\\cline{1-4}\end{array}$

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

Therefore:

$\textsf{P(Route B and delayed)}=\dfrac{8}{110}$

$\textsf{P(Route A and on time)}=\dfrac{28}{110}$

$\textsf{P(Route C and on time)}=\dfrac{24}{110}$

$\textsf{P(Route C and delayed)}=\dfrac{6}{110}$

So the event that has the highest probability is:

B.  The bus is from route A and is on time.