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Leila

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j_shreyans Legendary Member Default Avatar
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Leila

Post Wed Oct 29, 2014 10:03 am
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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Post Thu Jun 08, 2017 5:04 pm
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)⁴ + (1/5)⁴ * 4² =>

(1/5)⁴ * (1 + 16) =>

17/625

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Post Sat Jun 10, 2017 7:36 am
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
We need to determine the probability that Leila succeeds on exactly 3 throws or on all 4 throws.

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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Vellyanova Newbie | Next Rank: 10 Posts
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Post Wed Jun 07, 2017 11:53 pm
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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binit Master | Next Rank: 500 Posts Default Avatar
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Post Sun May 03, 2015 3:19 am
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.

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GMATinsight Legendary Member
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Post Fri Nov 28, 2014 7:11 am
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
Chances of Success = 1/5
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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Mathsbuddy Master | Next Rank: 500 Posts Default Avatar
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Post Tue Nov 11, 2014 5:43 am
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

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GMAT/MBA Expert

Post Wed Oct 29, 2014 11:00 am
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)1/5⁴

B)1/5³

C)6/5⁴

D)13/5⁴

E)17/5⁴

OAE
Give: P(succeeds on 1 throw) = 1/5

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
= 1/5⁴

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
= 4/5⁴
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) = 4/5⁴ + 4/5⁴ + 4/5⁴ + 4/5⁴
= 16/5⁴

So, P(succeeds AT LEAST 3 times) = P(succeeds 4 times) + P(succeeds 3 times)
= 1/5⁴ + 16/5⁴
= 17/5⁴
= E


Cheers,
Brent

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