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03 Nov 2024, 03:03
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
38% (03:01) correct
62%
(02:58)
wrong
based on 68
sessions
History
Date
Time
Result
Not Attempted Yet
Paul participates in an archery competition. He can attempt either one shot or three shots in order to win a prize of $1,000,000. If he attempts one shot, then to win the prize, he has to hit the target in that shot. However, if he attempts three shots, then to win the prize, he has to hit the target at least twice in three shots. Paul occasionally hits the target and occasionally misses the target. The probability of Paul hitting the target in any shot is x. Would Paul have a better chance of winning the prize if he attempted one shot?
(1) x > 0.3
(2) x < 0.4
(1) x > 0.3
(2) x < 0.4
03 Nov 2024, 11:04
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P(Hit) = \(x\)
P(Miss) = \((1-x)\)
To win, Paul must hit the target.
Question
\(x > 3x^2 - 2x^3\)
\(2x^3 - 3x^2 + x > 0\)
\(x(2x^2 - 3x + 1) > 0\)
\(x(2x^2 - 2x - x + 1) > 0\)
\(x(2x(x-1) - 1 (x - 1)) > 0\)
\(x(2x-1)(x-1) > 0\)
The critical points are
x = 0
x = 1
x = 1/2
---- -ve -- 0 --- + ve --- 1/2 --- -ve -- 1 ---- + ve --
As the value of x must be between 0 and 1, the question essentially translates to
Is 0 < x < 1/2 ?
Statement 1
(1) x > 0.3
While we know that x > 0.3, we can't for sure say that x < 0.5
For example,
Hence, we can eliminate option A, and D
Statement 2
(2) x < 0.4
This statement is sufficient. If x < 0.4, the value of x will always lie between 0 and 0.5
Option B
P(Miss) = \((1-x)\)
To win, Paul must hit the target.
- Probability of Paul hitting in one shot ⇒
- Paul gets one chance, and he must hit the target in that chance.
Net Probability = \(x\)
- Paul gets one chance, and he must hit the target in that chance.
- Probability of Paul hitting in three shot ⇒
- Paul gets three chances, and to win he must hit at least two of the three chances.
- Case 1: Paul hits twice and misses one chance.
Required Probability = \(x * x * (1-x) * \frac{3!}{2!}\) - Case 2: Paul hits all three shots
Required Probability = \(x * x * x\)
- Case 1: Paul hits twice and misses one chance.
- Paul gets three chances, and to win he must hit at least two of the three chances.
Question
\(x > 3x^2 - 2x^3\)
\(2x^3 - 3x^2 + x > 0\)
\(x(2x^2 - 3x + 1) > 0\)
\(x(2x^2 - 2x - x + 1) > 0\)
\(x(2x(x-1) - 1 (x - 1)) > 0\)
\(x(2x-1)(x-1) > 0\)
The critical points are
x = 0
x = 1
x = 1/2
---- -ve -- 0 --- + ve --- 1/2 --- -ve -- 1 ---- + ve --
As the value of x must be between 0 and 1, the question essentially translates to
Is 0 < x < 1/2 ?
Statement 1
(1) x > 0.3
While we know that x > 0.3, we can't for sure say that x < 0.5
For example,
- If x = 0.4, the response to the question Is 0 < x < 1/2 is Yes
- If x = 0.6, The response to the question Is 0 < x < 1/2 is No
Hence, we can eliminate option A, and D
Statement 2
(2) x < 0.4
This statement is sufficient. If x < 0.4, the value of x will always lie between 0 and 0.5
Option B
03 Nov 2024, 13:24
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siddhantvarma
(1) x > 0.3
Lets take two extreme cases to check. Let p be \(.4\), then:
Is .\(4 > .4*.4*.6 *3 + .4*.4*.4 \)
Is \(.4 > .352 -\) Yes
Let \(p\) be \(.9\)
\(.9 > .9*.9*.1 *3 + .9*.9*.9\)
\(.9 > .972\) - No
INSUFF.
(2) x < 0.4
\(.3 > .3*.3*.7 *3 + .3*.3*.3 \)
\(.3 > .189- \)Yes
Taking one shot will always be better.
SUFF.
Ans B
Note: It is not necessary to check for p =.2 , .1 etc because for positive numbers less than one, the more such numbers are multiplied the smaller they become. Hence RHS will always be smaller than LHS. for statement (2).
Hope it helped.
