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Leila

This topic has 3 expert replies and 4 member replies
j_shreyans Legendary Member Default Avatar
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Leila

Post Wed Oct 29, 2014 10:03 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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 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


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    Brent

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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

    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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    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.

    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

    GMAT/MBA Expert

    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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