Leila's carnival game
This topic has expert replies
- himu
- Master | Next Rank: 500 Posts
- Posts: 345
- Joined: Sun Dec 21, 2008 10:57 pm
- Thanked: 6 times
- Followed by:3 members
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?
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
P(at least 3 throws) = P(exactly 3 throws) + P(all 4 throws).himu 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?
Let W = win and L = lose.
Since P(W) = 1/5, P(L) = 4/5.
Case 1: Leila wins on exactly 3 throws
One way to win on exactly 3 throws is to lose only on the first throw:
P(LWWW) = 1/5 * 1/5 * 1/5 * 4/5 = 4/5�.
This result represents ONE WAY to win on exactly 3 throws.
Now we must account for ALL OF THE WAYS to win on exactly 3 throws.
Since L could be the 1st, 2nd, 3rd, or 4th throw -- for a total of FOUR WAYS -- we multiply by 4:
4 * 4/5� = 16/5�.
Case 2: Leila wins on all 4 throws
P(WWWW) = 1/5 * 1/5 * 1/5 * 1/5 = 1/5�.
Since either Case 1 OR Case 2 will yield a favorable outcome, we ADD the probabilities:
16/5� + 1/5� = 17/5�.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7312
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
We need to determine the probability that Leila succeeds on exactly 3 throws or on all 4 throws.himu 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?
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.
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Given: P(succeeds on 1 throw) = 1/5Leila 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