sandy wrote:

If the probability that the first event will occur is \(\frac{1}{4}\), and the probability that the second event will occur is \(1 \div \sqrt{x+2}\), then what is the probability that both events will occur?

A. \(\sqrt{x+2} \div (4x + 8)\)

B. \(\sqrt{x+2} \div 4\)

C. \(\sqrt{x+2} \div (16x + 32)\)

D. \(4 \div \sqrt{x+2}\)

E. \(4 \times \sqrt{x+2}\)

Drill 2

Question: 3

Page: 524

The event occuring together P(A and B) = P(A) * P (B)

here we write as p(\(\frac{1}{4}\) and \(1 \div \sqrt{x+2}\) ) = \(\frac{1}{4}\) * \(\frac{1}{{\sqrt{x+2}}}\)

or we can write as = \(\frac{1}{4}\) * \(\frac{1}{{\sqrt{x+2}}}\) * \({\sqrt{x+2}}\div{\sqrt{x+2}}\)

or we can write as = \(\sqrt{x+2} \div (4x + 8)\) (since 4 * \(\sqrt{x+2} * \sqrt{x+2}\) = 4x + 8 )

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