I believe that you must multiply the probability of getting bifocals (7/50) by itself to get the prob of picking two. (7/50)(7/50)
Same idea for single vision correction but use (43/50)(43/50)
50 employees in an office wear eyeglasses.
43
have single-vision correction, and
7
wear bifocals. If two employees are selected at random from this group, what is the probability that both of them wear bifocals? What is the probability that both have single-vision correction?
Round your answers to four decimal places.
4 answers
there are 50C2 ways of selecting 2 people from 50
bifocals ... there are 7C2 ways of selecting 2 people from 7
... probability is ... (7C2) / (50C2)
single-vision ... there are 43C2 ways of selecting 2 people from 43
... probability is ... (43C2) / (50C2)
bifocals ... there are 7C2 ways of selecting 2 people from 7
... probability is ... (7C2) / (50C2)
single-vision ... there are 43C2 ways of selecting 2 people from 43
... probability is ... (43C2) / (50C2)
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40 employees in an office wear eyeglasses. 23 have single-vision correction, and 17 wear bifocals. If two employees are selected at random from this group, what is the probability that both of them wear bifocals? What is the probability that both have single-vision correction?