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olchik [2.2K]
4 years ago
11

Name the values of the given digits the 4s in a 4400

Mathematics
1 answer:
IRISSAK [1]4 years ago
5 0
The values of 4s that are present in 4400 are 4 thousands and 4 hundreds.
4400 = 4 thousands + 4 hundreds + 0 tens + 0 ones
so the value of first 4 is 4 thousands and second 4 is 4 hundreds.
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We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

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So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

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So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

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