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If the probability that the first event will occur is , and
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24 Nov 2017, 20:52
24 Nov 2017, 20:52
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Expert Reply
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Question Stats:
76% (01:31) correct
23% (01:13) wrong based on 56 sessions
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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. \(4 \times \sqrt{x+2}\)
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
Question: 3
Page: 524
Re: If the probability that the first event will occur is , and
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02 Dec 2017, 07:36
02 Dec 2017, 07:36
2
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. \(4 \times \sqrt{x+2}\)
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
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 )
Re: If the probability that the first event will occur is , and
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29 Dec 2017, 13:17
29 Dec 2017, 13:17
Expert Reply
Explanation
Plug In to make this problem much simpler. If you plug in x = 2, then the probability for the second event is: \(1 \div \sqrt{4}=\frac{1}{2}\).
Now, because this is an “and” probability problem, you multiply the two probabilities together to find the target answer:
\(\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}\).
Choice A is the only one that works.
Plug In to make this problem much simpler. If you plug in x = 2, then the probability for the second event is: \(1 \div \sqrt{4}=\frac{1}{2}\).
Now, because this is an “and” probability problem, you multiply the two probabilities together to find the target answer:
\(\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}\).
Choice A is the only one that works.
