Respuesta :
Answer:
Therefore the only statement that is not true is b.)
Step-by-step explanation:
There employees are 6 secretaries, 5 consultants and 4 partners in the firm.
a.) The probability that a secretary wins in the first draw
= [tex]\frac{number \hspace{0.1cm} of \hspace{0.1cm} secreataries}{total \hspace{0.1cm} number \hspace{0.1cm} of \hspace{0.1cm} employees} = \frac{6}{15}[/tex]
b.) The probability that a secretary wins a ticket on second draw. It has been given that a ticket was won on the first draw by a consultant.
p(secretary wins on second draw | consultant wins on first draw)
=[tex]\frac{p((consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)\cap( secretary\hspace{0.1cm} wins\hspace{0.1cm} on second \hspace{0.1cm}draw))}{p(consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)}[/tex]
= [tex]\frac{\frac{5}{15} \times \frac{6}{14}}{\frac{5}{15} } = \frac{6}{14}[/tex] .
The probability that a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw = [tex]\frac{6}{15}[/tex] is not true.
The probability that a secretary wins on the second draw = [tex]\frac{number \hspace{0.1cm} of \hspace{0.1cm} secretaries \hspace{0.1cm} remaining } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} remaining} = \frac{6 - 1}{15 - 1} = \frac{5}{14}[/tex]
c.) The probability that a consultant wins on the first draw =
[tex]\frac{number \hspace{0.1cm} of \hspace{0.1cm} consultants \hspace{0.1cm} } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} } = \frac{5 }{15} = \frac{1}{3}[/tex]
d.) The probability of two secretaries winning both tickets
= (probability of a secretary winning in the first draw) × (The probability that a secretary wins on the second draw)
= [tex]\frac{6}{15} \times \frac{5}{14} = \frac{1}{7}[/tex]
Therefore the only statement that is not true is b.)
