Question

Alexis wanted to understand whether grade level had any relationship to their opinion on extending the school day. She surveyed some students and displayed the results in the table below:
In favor Opposed Undecided
Grade 9 6 4 10
Grade 10 9 13 10
Grade 11 16 17 11
Grade 12 13 7 12


Compare P(Grade 10 | opposed) with P(opposed | Grade 10).
P(Grade 10 | opposed) = P(opposed | Grade 10)
P(Grade 10 | opposed) < P(opposed | Grade 10)
P(Grade 10| opposed) > P(opposed | Grade 10)
There is not enough information.

PLEASE someone help me. Thank you.

Answer 1

Actually there is enough information to solve this problem. First, let us find the total per row and per column.

 (see attached pic)

P(Grade 10 | opposed) with P(opposed | Grade 10)

P(Grade 10 | opposed) = Number in Grade 10 who are opposed / Total number of Opposed (column)

P(Grade 10 | opposed) = 13 / 41 = 0.3171

 

P(opposed | Grade 10) = Number in Grade 10 who are opposed / Total number in Grade 10 (row)

P(opposed | Grade 10) = 13 / 32 = 0.4063

 

Therefore:

P(Grade 10 | opposed) IS NOT EQUAL P(opposed | Grade 10), hence they are dependent events.

 

Answer:

P(Grade 10 | opposed) < P(opposed | Grade 10)

Answer 2

Answer:

B. P(Grade 10|Opposed) < P(Opposed|Grade 10)

Step-by-step explanation:

We are given,

The table representing the relation between grade level and extending school days.

The conditional probability of A given that B has occurred is,

$P(A|B)=\frac{P(A\bigcap B)}{P(B)}$.

So, we have,

$P(Grade 10|Opposed)=\frac{P(Grade10\bigcap Opposed)}{P(Opposed)}$

i.e. $P(Grade 10|Opposed)=\frac{13}{41}$

i.e. P(Grade 10|Opposed) = 0.32

Also, we have,

$P(Opposed|Grade10)=\frac{P(Opposed\bigcap Grade10)}{P(Grade10)}$

i.e. $P(Opposed|Grade10)=\frac{13}{32}$

i.e. P(Opposed|Grade10) = 0.41

Thus, we see that,

0.32 = P(Grade 10|Opposed) < P(Opposed|Grade 10) = 0.41

Hence, option B is correct.