Question

A small-time bingo card costs P100.00 for 5 games. The prize for the first three games is P5,000.00, the fourth is P10,000.00 and the last prize is P20,000.00. If 1,000 bingo cards are going to be sold and you could only win once, what is the expected value of a ticket?

Answer 1

Answer: P91

Step-by-step explanation:

P5000.00 :

$p_{1} = \frac{1}{1000} . \frac{3}{5 } = \frac{3}{5000}$

Probability of the prizeP10,000.00 :

$p_{2} = \frac{1}{1000} . \frac{1}{5} = \frac{1}{5000}$

Probability of the prize P20,000.00:

$p_{3} = \frac{1}{1000} . \frac{1}{5} = \frac{1}{5000}$

The expected value of a ticket :

E(X) = 100 * 0.999 + (100 - 5000)* 3/5000 + (100 - 20000)*1/5000

= 99.9 - 44500/5000

=99.9 - 8.9 = $P_{91}$

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Answer 2

Answer:

The expected value of a ticket is $P_{91}$

Step-by-step explanation:

TO FIND: The expected value of a ticket is

GIVEN:

A small-time bingo card costs P100.00 for 5 games.

The prize for the first three games is P5,000.00.

 The probability of the first three games of P5000 is

    $P_{1} =\frac{1}{1000} *\frac{3}{5} =\frac{3}{5000}$

The fourth is  P10,000.00 .

The probability of fourth prize of P10000 is

   $P_{2} =\frac{1}{1000} *\frac{1}{5} =\frac{1}{5000}$

The last prize is P20,000.00.

The probability of last prize of P20000 is

$P_{3} =\frac{1}{1000} *\frac{1}{5} =\frac{1}{5000}$

If 1,000 bingo cards are going to be sold and you could only win once ,Then  the expected value of a ticket

               $E(X)=100*0.999+(100-5000)*\frac{3}{5000} +(100-20000)*\frac{1}{5000}$

               $E(X)=99.9-\frac{44500}{5000}$

               $E(X)=99.9-8.9$

                $E(X)=91$

                 $E(X)=P_{91}$

       

The expected value of a ticket is $P_{91}$

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