In a group of 25 students, 6 study both Art and Biology. 10 study Biology but not Art. 3 study neither subject. Given that a ran
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Let's say Bailey takes "b" days to paint it.
and Andrew takes "a" days to paint the same house.
now, Andrew is 6 times faster than Bailey, therefore, if Andrew takes "a" days to do it, Bailey takes then "6a" days, or b = 6a.
now, the year they worked together, they finished it in 7 days.
so, after 1 day then, they have only done 1/7 of the whole work.
and Andrew for one day, has done 1/a of the house, whilst Bailey has done 1/b of the house or 1/(6a).
and Andrew takes "a" days to paint the same house.
now, Andrew is 6 times faster than Bailey, therefore, if Andrew takes "a" days to do it, Bailey takes then "6a" days, or b = 6a.
now, the year they worked together, they finished it in 7 days.
so, after 1 day then, they have only done 1/7 of the whole work.
and Andrew for one day, has done 1/a of the house, whilst Bailey has done 1/b of the house or 1/(6a).
Answer:
P(a junior or a senior)=1
Step-by-step explanation:
The formula of the probability is given by:
Where P(A) is the probability of occurring an event A, n(A) is the number of favorable outcomes and N is the total number of outcomes.
In this case, N is the total number of the students of statistics class.
N=18+10=28
The probability of the union of two mutually exclusive events is given by:
Therefore:
P(a junior or a senior) =P(a junior)+P(a senior)
Because a student is a junior or a senior, not both.
n(a junior)=18
n(a senior)=10
P(a junior)=18/28
P(a senior) = 10/28
P(a junior or a senior) = 18/28 + 10/28
Solving the sum of the fractions:
P(a junior or a senior) = 28/28 = 1