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N76 [4]
2 years ago
15

A) what is the probability that the first contestant is a boy?

Mathematics
1 answer:
tamaranim1 [39]2 years ago
7 0

Probabilities are used to determine the chances of an event.

  • The probability that the first contestant is a boy is 1/2
  • The probability that the second contestant is a boy, if the first is a boy is 9/19
  • The probabilities are not independent

The given parameters are:

\mathbf{Contestants= 20}

\mathbf{Girls= 10}

\mathbf{Boys = 10}

<u>(a) Probability that first contestant is a boy</u>

This is calculated as:

\mathbf{P(Boy) = \frac{Boys}{Contestants}}

This gives

\mathbf{P(Boy) = \frac{10}{20}}

\mathbf{P(Boy) = \frac{1}{2}}

<u>(b) Probability that the second contestant is a boy, if the first is a boy</u>

This is calculated as:

\mathbf{P(Boy) = \frac{Boys - 1}{Contestants - 1}}

We subtracted 1, because the first contestant (a boy) is no longer part of the selection.

So, we have:

\mathbf{P(Boy) = \frac{10-1}{20-1}}

\mathbf{P(Boy) = \frac{9}{19}}

<u>(c) Independent probabilities</u>

The probabilities are not independent, because when a contestant is selected, the number of contestant is reduced by 1.

And this will affect the probability of the next selection.

Read more about probabilities at:

brainly.com/question/11234923

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Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!
LUCKY_DIMON [66]

Answer:

=8\sqrt{15}b^{\frac{7}{2}}

Step-by-step explanation:

\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}

=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}

=\sqrt{40}b^2\sqrt{24b^3}

\sqrt{24b^3}

=\sqrt{24}\sqrt{b^3}

=\sqrt{24}b^{\frac{3}{2}}

=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

\sqrt{2^3}

=2^{3\cdot \frac{1}{2}

=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}

2^{\frac{3}{2}+\frac{3}{2}}

=2^3

=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}

b^{\frac{3}{2}+2}

=b^{\frac{7}{2}}

=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}

=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}

=8\sqrt{15}b^{\frac{7}{2}}

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