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06 May 2021, 01:30
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E
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Difficulty:
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Question Stats:
48% (02:39) correct
52%
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wrong
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A game at the state fair has a circular target with a radius of 10 cm on a square board measuring 30 cm on a side. Players win prizes if they throw two darts and hit only the circular area on at least one of the two attempts. What is the probability that Jim won the game?
(A) \(1-\frac{9-π}{9}\)
(B) \(1-\frac{18π-π^2}{81}\)
(C) \(\frac{9-π}{9}\)
(D) \(\frac{(9-π)^2}{81}\)
(E) \(\frac{18π-π^2}{81}\)
(A) \(1-\frac{9-π}{9}\)
(B) \(1-\frac{18π-π^2}{81}\)
(C) \(\frac{9-π}{9}\)
(D) \(\frac{(9-π)^2}{81}\)
(E) \(\frac{18π-π^2}{81}\)
07 May 2021, 01:15
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Bunuel
A game at the state fair has a circular target with a radius of 10 cm on a square board measuring 30 cm on a side. Players win prizes if they throw two darts and hit only the circular area on at least one of the two attempts. What is the probability that Jim won the game?
(A) \(1-\frac{9-π}{9}\)
(B) \(1-\frac{18π-π^2}{81}\)
(C) \(\frac{9-π}{9}\)
(D) \(\frac{(9-π)^2}{81}\)
(E) \(\frac{18π-π^2}{81}\)
(A) \(1-\frac{9-π}{9}\)
(B) \(1-\frac{18π-π^2}{81}\)
(C) \(\frac{9-π}{9}\)
(D) \(\frac{(9-π)^2}{81}\)
(E) \(\frac{18π-π^2}{81}\)
The area of the circle is \(\pi{r^2}=100\pi\);
The area of the square is \(side^2=900\).
The probability of hitting the circle is therefore \(\frac{100\pi}{900}=\frac{\pi}{9}\) and the probability of missing the circle is \(1-\frac{\pi}{9}=\frac{9-\pi}{9}\).
The probability of hitting the circle on at least one of the two attempts = 1 - {the probability of missing on both of the two attempts} =
\(1-\frac{9-\pi}{9}*\frac{9-\pi}{9}=\frac{81-(9-\pi)^2}{81}=\frac{81-(81-18\pi+{\pi}^2)}{81}=\frac{18\pi-{\pi}^2}{81}\).
Answer: E.
Hope it's clear.
General Discussion
06 May 2021, 02:19
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A game at the state fair has a circular target with a radius of 10 cm on a square board measuring 30 cm on a side. Players win prizes if they throw two darts and hit only the circular area on at least one of the two attempts. What is the probability that Jim won the game?
Area of the square = \(30^2\) = 900 \(cm^2\)
Area f circle = π*\(10^2\) = 100π \(cm^2\)
Missed area for the target = 900- 100π
P (Both throws missed the target) = \(\frac{(900- 100π)}{900}\) *\(\frac{(900- 100π)}{900}\)
P(atleast one hit the target) = 1- \(\frac{(900- 100π)}{900}\) *\(\frac{(900- 100π)}{900}\) = \(\frac{(81- 81-π^2 +18π)}{81}\) = \(\frac{(18π-π^2)}{81}\)
So, I think E.
Area of the square = \(30^2\) = 900 \(cm^2\)
Area f circle = π*\(10^2\) = 100π \(cm^2\)
Missed area for the target = 900- 100π
P (Both throws missed the target) = \(\frac{(900- 100π)}{900}\) *\(\frac{(900- 100π)}{900}\)
P(atleast one hit the target) = 1- \(\frac{(900- 100π)}{900}\) *\(\frac{(900- 100π)}{900}\) = \(\frac{(81- 81-π^2 +18π)}{81}\) = \(\frac{(18π-π^2)}{81}\)
So, I think E.
