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In a statistics class there are 11 juniors and 6 seniors; 2 of the seniors are females; 6 of the juniors are males. if a student is selected at random, what is the probability that the student is either a junior or a female?

enter your answer as a fully reduced fraction.

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hankaF
The probability is 8/17.

The tricky portion of this question is the phrasing using the word "or".
The fact that it is looking for either a female "or" a junior, let's us know that the selection can't be both. In mathematics, we would need the term "and/or" in order to include those that are both.
So, given that information, we know we have 11 juniors, 6 of which are males and 5 of which are females. We would include the males in our selection.
We also know there are 6 seniors, 2 of which are female and 4 of which are males. We would include the 2 females.

This would give us a selection size of 8. There are 17 total kids, so we divide the 8 by 17 for the answer of 8/17.
JeanaShupp

Answer: [tex]\dfrac{13}{17}[/tex]

Step-by-step explanation:

Given: Number of juniors = n(J)= 11

Number of seniors = 6

Total students n(S)=[tex]11+6=17[/tex]

Number of seniors are which are females =2

Number of  juniors are which are males =6

Then, number of juniors which are females = [tex]11-6=5[/tex]

Now, total females n(F)=[tex]2+5=7[/tex]

such that number of juniors which are females n(J∩F) = [tex]5[/tex]

Now, the number of students either a junior or a female is given by :_

[tex]n(J\cup F)=n(J)+n(F)-n(J\cap F)\\\\\Rightarrow n(J\cup F)=11+7-5=13[/tex]

Now, the probability that the student is either a junior or a female is given by :-

[tex]P(J\cup F)=\dfrac{n(J\cup F)}{n(S)}=\dfrac{13}{17}[/tex]

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