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3 years ago

Which pair is made up of prime numbers 11 & 33 , 42 & 56 , 37 & 67, 57 & 97

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

3 0

Answer:

37 & 67

Step-by-step explanation:

11 & 33

11 - 1, 11

11 is prime

33 - 1, 3, 11, 33

33 is composite number

11 & 33 is incorrect

42 & 56

42 - 1, 2, 3, 6, 7, 14, 21, 42
56 - 1, 2, 4, 7, 8, 14, 28, 56
Both 42 and 56 are composite numbers
42 & 56 is incorrect

37 & 67

37 - 1, 37
67 - 1, 67
Both 37 and 67 are prime numbers
37 & 56 is correct

57 & 97

57 - 1, 3, 19, 57
97 - 1, 97
57 is a composite number, 97 is a prime number
57 & 97 is incorrect

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B) 0.283

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

25% of the students drive themselves to school.

This means that $p = 0.25$

Class of 18 students

This means that $n = 18$

What would be the probability that at least 6 students drive themselves to school?

This is

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{18} = 0.006$

$P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{17} = 0.034$

$P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{16} = 0.096$

$P(X = 3) = C_{15,3}.(0.25)^{3}.(0.75)^{15} = 0.17$

$P(X = 4) = C_{15,4}.(0.25)^{4}.(0.75)^{14} = 0.213$

$P(X = 5) = C_{15,5}.(0.25)^{5}.(0.75)^{13} = 0.199$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.006 + 0.034 + 0.096 + 0.17 + 0.213 + 0.199 = 0.718$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.718 = 0.282$

Closest option is B, just a small rounding difference.

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