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02 Nov 2020, 08:30
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A class consists of 10 boys and 20 girls. Exactly half of the boys and half of the girls have brown eyes. Determine the probability that a randomly selected student will be a boy or a student with brown eyes.
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{2}{3}\)
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{2}{3}\)
02 Nov 2020, 08:39
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𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 𝑎𝑛𝑑 𝐵)
P(A) = random select a boy = 10/30= 1/3
P(B) = random select a brown eye = half of the class = 15/30=1/2
P( A and B) = boys with brown eyes/entire class = 5/30 = 1/6
1/3 + 1/2 - 1/6 = 2/3
P(A) = random select a boy = 10/30= 1/3
P(B) = random select a brown eye = half of the class = 15/30=1/2
P( A and B) = boys with brown eyes/entire class = 5/30 = 1/6
1/3 + 1/2 - 1/6 = 2/3
02 Nov 2020, 17:28
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The main learning here is that a boy is not attached to brown eyes.
Here we have to calculate the probability of selecting a boy, at first, = Number of boys/total students = 10/30= 1/3
A student with brown eyes = (5+10)/30= 1/2
A boy with brown eyes = 5/30= 1/6
SO going with "or" formula = 1/3 + 1/2 -1/6 = 2/3 Answer.
Here we have to calculate the probability of selecting a boy, at first, = Number of boys/total students = 10/30= 1/3
A student with brown eyes = (5+10)/30= 1/2
A boy with brown eyes = 5/30= 1/6
SO going with "or" formula = 1/3 + 1/2 -1/6 = 2/3 Answer.
