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Anna007 [38]
2 years ago
13

A carnival game allows a group of players to each draw and keep a marble from a bag. The bag contains 5 gold marbles, 25 silver

marbles, and 70 red marbles. A player wins a large prize for drawing a gold marble and a small prize for drawing a silver marble. There is no prize for drawing a red marble. At the start of the game, the probability of winning a large prize is 0.05 and the probability of winning a small prize is 0.25. Suppose that the first player draws a silver marble and wins a small prize. What is the probability that the second player will also win a small prize? If a group of four plays the game one at a time and everyone wins a small prize, which player had the greatest probability of winning a large prize? How could the game be made fair for each player? That is, how could you change the game so that each player has an equal chance of winning a prize?
Mathematics
2 answers:
Step2247 [10]2 years ago
8 0
The probability is lower for the second player to win a prize. The first player has a bigger probability of winning a large prize. To make the game fair for each player, each player would have to choose a marble from the bag at the same time.
Nezavi [6.7K]2 years ago
4 0

1. 25% too because it wont change if they have the same amount of marbles.

2. 5% since there is very little gold marbles.

3. Take out 55 red marbles so it could be fair or just take out red marbles in general. That would be fair.

Hope I helped. Have a great day

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Answer:

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Austin keeps a right conical basin for the birds in his garden as represented in the diagram. The basin is 40 centimeters deep,
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There are 10 pots exposed in the shop, 2 of which have hidden defects. the customer buys two pieces. what is the probability tha
Tpy6a [65]
Check the tree diagram, of all the possible scenarios and the probabilities.

All the possible scenarios are: 

1st.non defective,  2nd.non defective
1st.non defective,  2nd. defective
1st. defective,  2nd. non defective
1st. defective,  2nd. defective

consider the case 1st.non defective,  2nd.non defective

the probability that the first one is non defective is 8/10, since 8 out of 10 are non defective.

In this scenario, the second one is non defective as well. The probability for this to happen is 7/9, since now we have 7 non defective pots, and 9 in total.

This means that the probability of the first to be non defective, and the second non-defective is (8/10) * (7/9)=0.622

similarly we calculate the other cases, as shown in the picture.

P(at least one is defective)=
P(1st.non defective,  2nd. defective)+P(1st. defective,  2nd. non defective)+P(1st. defective,  2nd. defective)=
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Answer: 0.378

8 0
1 year ago
A bar of soap with dimensions 2 x 4 x 0.5 inches is sold for $3 individually. A three-pack of the same soap costs $8. How much d
LUCKY_DIMON [66]
We know that
volume of a bar of soap=2*4*0.5----> 4 in³

<span>A bar of soap is sold for $3 individually
</span>so
$3/4-----> 0.75 $/in³

<span>A three-pack of the same soap costs $8
so
$8*(3*4)----> 0.67 $/in</span>³

<span>if you buy the three-pack you save-----> [0.75-0.67]=0.08 $/in</span>³

the answer is
$0.08
7 0
2 years ago
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