Answer to Question #345723 in Statistics and Probability for Ttypre

Question

1. In a viral pole test it is known that in a group of five (5) people, exactly one (1) well test positive. If they are tested one by one in random under for confirmation, what is the probability that only two (2) tests are needed?

2. A basket of fruits contains eight (8) apples and ten (10) oranges. Half of the apples and half of the oranges are rotten. If one (1) fruit is chosen at random, what is the probability that a rotten apple or an orange is chosen?

3. A small-time bingo card cost 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?

7. You pick a card from a deck. If it is a face card, you will win P500.00. if you get an ace, you will win P1,000. If the card you picked is red you get to P100.00. for any other card, you will win nothing. Find the expected value that you can possibly win.

Answer 1

1.

$$P(1^{st}Negative)=\dfrac{5-1}{5}$$

$$P(2^{nd}Positive)=\dfrac{1}{5-1}$$

$$P(X=2)=P(1^{st}Negative\cap2^{nd}Positive)$$

$$=P(1^{st}Negative)P(2^{nd}Positive)$$

$$=\dfrac{5-1}{5}(\dfrac{1}{5-1})=\dfrac{1}{5}$$

2.

There are $8+10=18$ fruits in the basket. There are $8/2+10/2$ rotten fruits.

$$P(rotten\ apple\ or\ rotten\ orange)$$

$$=P(rotten\ fruit)$$

$$=\dfrac{8/2+10/2}{8+10}=\dfrac{9}{18}=\dfrac{1}{2}$$

$$P(rotten\ apple\ or\ any\ orange)$$

$$=P(rotten\ apple\cup any\ orange)$$

$$=\dfrac{8/2+10}{8+10}=\dfrac{14}{18}=\dfrac{7}{9}$$

3. Probability of the prize P5,000.00:

$$p_1=\frac{1}{1000}\cdot\frac{3}{5}=\frac{3}{5000}$$

Probability of the prize P10,000.00:

$$p_2=\frac{1}{1000}\cdot\frac{1}{5}=\frac{1}{5000}$$

Probability of the prize P20,000.00:

$$p_3=\frac{1}{1000}\cdot\frac{1}{5}=\frac{1}{5000}$$

The expected value of a ticket:

$$E(X)=-100(\frac{1000-1}{1000})+(5000-100)(\frac{3}{5000})$$

$$+(10000-100)(\frac{1}{5000})+(20000-100)(\frac{1}{5000})$$

$$=-91$$

7. There are 52 cards in the deck.

There are 12 face cards.

There are 4 aces.

There are 26 red cards.

$$E(X)=\dfrac{12}{52}(500)+\dfrac{4}{52}(1000)+\dfrac{26}{52}(100)=242.31$$