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Shtirlitz [24]
3 years ago
10

A consulting firm has received 2 Super Bowl playoff tickets from one of its clients. To be fair, the firm is randomly selecting

two different employee names to 'win' the tickets. There are 6 secretaries, 5 consultants and 4 partners in the firm. Which of the following statements is not true?
The probability of a secretary winning a ticket on the first draw is 6/15.
The probability of a secretary winning a ticket on the second draw given a consultant won a ticket on the first draw is 6/15.
The probability of a consultant winning a ticket on the first draw is 1/3.
The probability of two secretaries winning both tickets is 1/7.
Mathematics
1 answer:
Soloha48 [4]3 years ago
5 0

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

= \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}

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)

=\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)}

= \frac{\frac{5}{15}  \times \frac{6}{14}}{\frac{5}{15} }  = \frac{6}{14}  .

The probability that  a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw  = \frac{6}{15} is not true.

The probability that a secretary wins on the second draw  = \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}

c.) The probability that a consultant wins on the first draw  =

\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}

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)

= \frac{6}{15}  \times \frac{5}{14}  = \frac{1}{7}

Therefore the only statement that is not true is b.)

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