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

The probability will be "10.83%".

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]* P[writes \ important \ notes]`

By putting the values, we get

=
`0.57* 0.19`

=
`0.1083`

=
`10.83` %