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Bunuel
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Bunuel
Ms Li works at an office where the work timing is from 9:00 AM to 6:00 PM. 25% of a year she goes late to office and 35% of a year she leaves early from office. If P is the probability that she works at office the entire working day then

A. 0.25 ≤ P ≤ 0.35
B. 0.25 ≤ P ≤ 0.65
C. 0.4 ≤ P ≤ 0.65
D. 0.35 ≤ P ≤ 0.4
E. 0.1 ≤ P ≤ 0.6
If Ms Li's late arrival probability does not overlaps with her early leaving probability then we are left with (0.25) + (0.35) + x = 1, where x is the probability(min.) of her working the entire day.
\(\implies x_{min} = 0.40\) [We can pick our answer at this stage]
If the two probabilities fully overlap then we have ((0.25) + 10) + x = 1, where now x is probability(max.) of her working the entire day.
\(\implies x_{max} = 0.65\)

Answer C.
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Can we do like this?

Not late= 0.75
Not leaving early= 0.65

Thus, full working day= not late percent of not leaving early i.e. 0.75*0.65 ( 75% of 65%)
= 0.4875 (48.75%)

Thus, option C.

Am I doing it right?
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ronaldoSuiiii
Can we do like this?

Not late= 0.75
Not leaving early= 0.65

Thus, full working day= not late percent of not leaving early i.e. 0.75*0.65 ( 75% of 65%)
= 0.4875 (48.75%)

Thus, option C.

Am I doing it right?
ronaldoSuiiii
Don't you think all the option fulfil the criteria you are applying. How only C is the answer??
P is some range which is what we need to establish. I guess you missed in that part.

Please check.
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unraveled
ronaldoSuiiii
Can we do like this?

Not late= 0.75
Not leaving early= 0.65

Thus, full working day= not late percent of not leaving early i.e. 0.75*0.65 ( 75% of 65%)
= 0.4875 (48.75%)

Thus, option C.

Am I doing it right?
ronaldoSuiiii
Don't you think all the option fulfil the criteria you are applying. How only C is the answer??
P is some range which is what we need to establish. I guess you missed in that part.


Thanks yeah.

Please check.
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