Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?
A)1/5^4
B)1/5^3
C)6/5^4
D)13/5^4
E)17/5^4
OAE
Leila
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Give: P(succeeds on 1 throw) = 1/5j_shreyans wrote:Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?
A)1/5�
B)1/5³
C)6/5�
D)13/5�
E)17/5�
OAE
P(succeeds at least 3 times) = P(succeeds 4 times OR succeeds 3 times)
= P(succeeds 4 times) + P(succeeds 3 times)
P(succeeds 4 times)
P(succeeds 4 times) = P(succeeds 1st time AND succeeds 2nd time AND succeeds 3rd time AND succeeds 4th time)
= P(succeeds 1st time) x P(succeeds 2nd time) x P(succeeds 3rd time) x P(succeeds 4th time)
= 1/5 x 1/5 x 1/5 x 1/5
= 1/5�
P(succeeds 3 times)
Let's examine one possible scenario in which Leila succeeds exactly 3 times:
P(FAILS the 1st time AND succeeds 2nd time AND succeeds 3rd time AND succeeds 4th time)
= P(FAILS the 1st time) x P(succeeds 2nd time) x P(succeeds 3rd time) x P(succeeds 4th time)
= 4/5 x 1/5 x 1/5 x 1/5
= 4/5�
Keep in mind that this is only ONE possible scenario in which Leila succeeds exactly 3 times (Leila fails the 1st time).
Leila can also FAIL the 2nd time, or the 3rd time or the 4th time.
Each of these probabilities will also equal 4/5³
So, P(succeeds 3 times) = 4/5� + 4/5� + 4/5� + 4/5�
= 16/5�
So, P(succeeds AT LEAST 3 times) = P(succeeds 4 times) + P(succeeds 3 times)
= 1/5� + 16/5�
= [spoiler]17/5�[/spoiler]
= E
Cheers,
Brent
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Possible throw combinations (using 0 = lose, 1 = win):
1111 -> p = (1/5)^4
1110 -> p = (1/5)^3 x 4/5 = 4(1/5)^4
1101 -> p = (1/5)^3 x 4/5 = 4(1/5)^4
1011 -> p = (1/5)^3 x 4/5 = 4(1/5)^4
0111 -> p = (1/5)^3 x 4/5 = 4(1/5)^4
Add them up:
TOTAL P = 17 x (1/5)^4
ANSWER E
1111 -> p = (1/5)^4
1110 -> p = (1/5)^3 x 4/5 = 4(1/5)^4
1101 -> p = (1/5)^3 x 4/5 = 4(1/5)^4
1011 -> p = (1/5)^3 x 4/5 = 4(1/5)^4
0111 -> p = (1/5)^3 x 4/5 = 4(1/5)^4
Add them up:
TOTAL P = 17 x (1/5)^4
ANSWER E
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Chances of Success = 1/5j_shreyans wrote:Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?
A)1/5^4
B)1/5^3
C)6/5^4
D)13/5^4
E)17/5^4
OAE
i.e.e Chances of Failure = 1-(1/5) = 4/5
Probability of Atleast 3 Successful includes
1) Probability of exactly 3 Successful attempt = 4C3 (1/5)^3 x (4/5)
2) Probability of exactly 4 Successful attempt = (1/5)^4
Total Probability = [4C3 (1/5)^3 x (4/5)] + [(1/5)^4] = (4x4 + 1) / (5^4) = 17/5^4
Answer: Option E
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P(wins atleast 3 times)= P(wins 3 times)+P(wins 4 times)
= No. of ways of selecting 3 of 4*P(wins)*P(wins)*P(wins)P(loses) + P(wins)P(wins)P(wins)P(wins)
= 4C3 * 1/5 */15 * 1/5 * 4/5 + 1/5 * 1/5 */15 * 1/5
= 16/5^4 + 1/5^4
= 17 / 5^4
~Binit.
= No. of ways of selecting 3 of 4*P(wins)*P(wins)*P(wins)P(loses) + P(wins)P(wins)P(wins)P(wins)
= 4C3 * 1/5 */15 * 1/5 * 4/5 + 1/5 * 1/5 */15 * 1/5
= 16/5^4 + 1/5^4
= 17 / 5^4
~Binit.
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j_shreyans wrote:Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?
A)1/5^4
B)1/5^3
C)6/5^4
D)13/5^4
E)17/5^4
OAE
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P(at least 3) = P(exactly 3) + P(exactly 4)
P(exactly 4) is easy => 1/5 * 1/5 * 1/5 * 1/5
P(exactly 3) is trickier, since we have four different arrangements to consider. (The miss could come on any of the four throws.) That means we've got four orders:
Hit, Hit, Hit, Miss
Hit, Hit, Miss, Hit
Hit, Miss, Hit, Hit
Miss, Hit, Hit, Hit
Each of these has the same probability (1/5 * 1/5 * 1/5 * 4/5), and since we've got four of them, we multiply that by 4:
P(exactly 3) = 1/5 * 1/5 * 1/5 * 4/5 * 4
Adding the two up, we're done!
(1/5)� + (1/5)� * 4² =>
(1/5)� * (1 + 16) =>
17/625
P(exactly 4) is easy => 1/5 * 1/5 * 1/5 * 1/5
P(exactly 3) is trickier, since we have four different arrangements to consider. (The miss could come on any of the four throws.) That means we've got four orders:
Hit, Hit, Hit, Miss
Hit, Hit, Miss, Hit
Hit, Miss, Hit, Hit
Miss, Hit, Hit, Hit
Each of these has the same probability (1/5 * 1/5 * 1/5 * 4/5), and since we've got four of them, we multiply that by 4:
P(exactly 3) = 1/5 * 1/5 * 1/5 * 4/5 * 4
Adding the two up, we're done!
(1/5)� + (1/5)� * 4² =>
(1/5)� * (1 + 16) =>
17/625
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We need to determine the probability that Leila succeeds on exactly 3 throws or on all 4 throws.j_shreyans wrote:Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop. If her chance of success on each throw is 1/5, what is the chance that she will succeed on at least 3 of the throws?
A)1/5^4
B)1/5^3
C)6/5^4
D)13/5^4
E)17/5^4
Scenario 1: succeeds on exactly 3 throws
We can let Y denote a successful throw and N denote a non-successful throw:
P(Y-Y-Y-N) = 1/5 x 1/5 x 1/5 x 4/5 = 4/(5^4)
However, we must account for the order of Y-Y-Y-N. Using our formula for indistinguishable items, Y, Y, Y, and N can be arranged in 4!/3! = 4 ways.
Thus, the probability of succeeding on exactly 3 throws (out of 4 attempts) is 4/(5^4) x 4 = 16/(5^4).
Now we can determine scenario 2:
Scenario 2: succeeds on all 4 attempts
P(Y-Y-Y-Y) = 1/5 x 1/5 x 1/5 x 1/5 = 1/(5^4)
Thus, the probability of succeeding on 3 throws out of 4 or 4 throws out of 4 is 16/(5^4) + 1/(5^4) = 17/5^4.
Answer: E
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