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Archy [21]
2 years ago
10

.1.In a large bag of marbles, 25% of them are red. A child chooses 4 marbles from this bag. If the child chooses the marbles at

random, what is the chance that the child gets exactly three red marbles? . A) 0.047. B) 0.141 . C) 0.211 . D) 0.063. . 2.A pet supplier has a stock of parakeets of which 10% are blue parakeets. A pet store orders 3 parakeets from this supplier. If the supplier selects the parakeets at random, what is the chance that the pet store gets exactly one blue parakeet?. A) 0.081 . B) 0.243 . C) 0.027 . D) 0.003 . In a large bag of marbles, 25% of them are red. A child chooses 4 marbles
Mathematics
2 answers:
professor190 [17]2 years ago
6 0
1. Your question tells that the percentage of the red marble in the bag is 25%. So if a child chooses 4 marbles, the probability that he could get 3 red marbles is A.0.047
2.In your question the supplier has a stock of parakeets were 10% is blue, so the chance of the supplier to deliver 1 blue parakeet out of 3 is letter B.0.243
WARRIOR [948]2 years ago
5 0

Answer:

A: 0.047

B: 0.243

Step-by-step explanation:

A:

\frac{4!}{3!(4-3)!} (0.25)^{3} (1-0.25)^{4-3}

= 0.0468 ≈ 0.047

B:

The probability of a blue parakeet to be chosen is 0.10 and the probability of a non blue parakeet is 0.90.

We can conclude from the given situation, the following combinations:

No blue parakeet:  0.90^{3}= 0.729

If 1 blue parakeet is selected: 0.90^{2}\times0.10= 0.081

When there are 2 blue parakeets: 0.90\times0.10^{2} =0.009

When 3 blue parakeets: 0.10^{3} =0.001

So,the chance that the pet store gets exactly one blue parakeet is given by:

0.081\times3 =0.243

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Take common denominator to combine fractions

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