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Register now and save up to $200 Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code ## Leila tagged by: Brent@GMATPrepNow This topic has 3 expert replies and 4 member replies j_shreyans Legendary Member Joined 07 Aug 2014 Posted: 510 messages Followed by: 5 members Thanked: 3 times #### Leila 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 Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums! ### GMAT/MBA Expert Brent@GMATPrepNow GMAT Instructor Joined 08 Dec 2008 Posted: 10862 messages Followed by: 1215 members Thanked: 5200 times GMAT Score: 770 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 _________________ Brent Hanneson – Founder of GMATPrepNow.com Use our video course along with Check out the online reviews of our course Come see all of our free resources Thanked by: j_shreyans GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! Mathsbuddy Master | Next Rank: 500 Posts Joined 08 Nov 2013 Posted: 447 messages Followed by: 1 members Thanked: 25 times 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 Joined 10 May 2014 Posted: 1001 messages Followed by: 21 members Thanked: 205 times 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 _________________ Prosper!!! Bhoopendra Singh & Sushma Jha "GMATinsight" Contact Us Testimonials To register for One-on-One FREE ONLINE DEMO Class Call/e-mail e-mail: info@GMATinsight.com Mobile: +91-9999687183 / +91-9891333772 Get in touch for SKYPE-Based Interactive Private Tutoring One-On-One Classes fee - US$40 per hour &
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binit Master | Next Rank: 500 Posts
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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.

Vellyanova Newbie | Next Rank: 10 Posts
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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

Matt@VeritasPrep GMAT Instructor
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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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Jeff@TargetTestPrep GMAT Instructor
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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.

_________________
Jeffrey Miller Head of GMAT Instruction

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