Question

According to a survey, 57% of all workers read e-mail while they are in a meeting. Suppose that 19% of those who read e-mail while they are in a meeting write down important notes. Assuming that these figures are true for all workers, if a worker is randomly selected, what is the probability the worker reads e-mail while in meeting and writes important notes

Answer 1

Answer:

The probability will be "10.83%".

Step-by-step explanation:

According to the question,

Workers read e-mail,

= 57%

= 0.57

Write down important notes,

= 19%

= 0.19

Now,

The probability of worker who reads e-mail while in meeting and writes important notes will be:

= $P[email \ reading \ during \ meeting]\times P[writes \ important \ notes]$

By putting the values, we get

= $0.57\times 0.19$

= $0.1083$

= $10.83$ %