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Leila is playing a carnival game in which she is given 4
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Updated on: 04 Oct 2012, 03:30

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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?

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04 Oct 2012, 03:38

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Topic moved to PS forum.

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

At least 3 out of 4 means 3 or 4, so \(P=C^3_4*(\frac{1}{5})^3*\frac{4}{5}+(\frac{1}{5})^4=\frac{17}{5^4}\).

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29 Aug 2016, 06:43

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Archit143 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

Given: 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 =

Re: Leila is playing a carnival game i
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03 Oct 2012, 22:29

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Archit143 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

pls explain the approach

Probability that She will succeed on at least 3 of the throws = Probability that she will succeed on exactly 3 throws + probability that she will succeed on atleast 4 throws.

Probability that she will succeed on exactly 3 throws: S - succeed F - Failure So there are 4 possibilities FSSS , SFSS, SSFS, SSSF

Probability os success = 1/5 Probability of failure= 1 - probability of success = 1- 1/5 = 4/5

Probability of success on at least three tries = Probability that she will succeed on exactly 3 throws + probability that she will succeed on atleast 4 throws = P(FSSS) + P(SFSS) + P(SSFS) + P(SSSF) + P(SSSS) = {(4/5) *(1/5) *(1/5) *(1/5)} + {(1/5) *(4/5) *(1/5) *(1/5)} + {(1/5) *(1/5) *(4/5) *(1/5)} + {(1/5) *(1/5) *(1/5) *(4/5)} + {(1/5) *(1/5) *(1/5) *(1/5)} = (1/5)*(1/5)*(1/5)* { (4/5) + (4/5) + (4/5) + (4/5) + (1/5) } = 17 / 5^4

So, answer will be E Hope it helps!
_________________

Re: Leila is playing a carnival game in which she is given 4
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16 Sep 2013, 02:52

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Phoenix72 wrote:

Bunuel,

I understood how this question is solved and why we multiple 4 *((1/5)^3 *(4/5)). However, i often get confused when this concept has to be used. E.g) in the below question, why is 3 not multipled to 7C2*3C1/10C2

A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?

First of all it should be \(\frac{C^2_7*C^1_3}{C^3_{10}}\), not 7C2*3C1/10C2.

Next, we should multiply if we use "simple probability" approach: \(P(BBR)=\frac{3!}{2!}*\frac{3}{10}*\frac{7}{9}*\frac{6}{8}\).

Re: Leila is playing a carnival game in which she is given 4
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22 Sep 2013, 05:47

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SaraLotfy wrote:

It seems that I am missing an important concept,

Can anyone please clarify why does the order matter here? How can I tell the difference between a normal probability question and a permutation that requires order.

The order matters because the question did not explicitly say that the 4th attempt will be the failure. It could be 1st,2nd,3rd or 4th throw that could result in failure and hence we add up those cases.

for each of them we have probability as 4/5^4 and for the case where all throws are successful=1/5^4. therefore total probability of the event= 4*4/5^4+1/5^4=17/5^4
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--It's one thing to get defeated, but another to accept it.

Re: Leila is playing a carnival game in which she is given 4
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17 Jun 2013, 22:25

pjreddy wrote:

Bunuel, I initially chose B for my answer. I solved the problem pretty much exactly the same way as you only I did not use the combinations formula when solving my answer. Can you explain why/how you used that in your explanation above?

i think you are forgetting to multiply the possibility of 3Success+1Failure by 4 because it would be SSSF or SSFS or SFSS or FSSS

Thus 4[4/625] and then adding that to probability of SSSS which is 1/625. thus 16/625+1/625=17/625

Re: Leila is playing a carnival game in which she is given 4
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16 Sep 2013, 02:23

Bunuel,

I understood how this question is solved and why we multiple 4 *((1/5)^3 *(4/5)). However, i often get confused when this concept has to be used. E.g) in the below question, why is 3 not multipled to 7C2*3C1/10C2

A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?

Re: Leila is playing a carnival game in which she is given 4
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18 Sep 2013, 07:29

It seems that I am missing an important concept,

Can anyone please clarify why does the order matter here? How can I tell the difference between a normal probability question and a permutation that requires order.

Probability - Why consider order while tossing a coin( or any event) ?
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30 Jun 2016, 11:53

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?

Explanation:

There are two ways that Leila can succeed on at least 3 of the throws—she can succeed on exactly 3 throws or on all 4. Find the total probability by determining the chance of each of these two distinct outcomes and adding them together.

Start with the chance of succeeding on all 4 throws. Since Leila’s chance of success is 1/5, multiply out her chance of succeeding 4 times in a row: 1/5 × 1/5 × 1/5 × 1/5 = 1/54

You can now eliminate answer A; it can’t be correct, because you haven’t yet added in the possibility of succeeding on exactly 3 throws.

If Leila succeeds on exactly 3 of the throws, then she must have 1 missed throw. The chance of missing a throw is 1 – 1/5 = 4/5, so multiply the individual probabilities to get the total probability: 1/5 × 1/5 × 1/5 × 4/5 = 4/54

However, this is the probability of only one specific outcome: hit, hit, hit, miss. Since she doesn’t need to make her throws in this specific order, you'll also need to consider the other possibilities: hit, hit, miss, hit hit, miss, hit, hit miss, hit, hit, hit I have a slight confusion in the highlighted section. I am not able to visualise why exactly do we need to consider the order when what I really want to know is HOW many times did she succeed. Can somebody help me understand the basic behind the "permutation" involved in events based questions and why dont we just consider them a "combination". I am sure its a very trivial question and mathematically we have always read that consider the order ( lets say of a flip of a coin) however I want to understand the underlying principle

Re: Leila is playing a carnival game in which she is given 4
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13 Jan 2018, 11:11

Archit143 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

at least 3 wins out of 4 chances can happen as follows

Re: Leila is playing a carnival game in which she is given 4
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18 Aug 2018, 02:13

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Archit143 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?

There are two ways that Leila can succeed on at least 3 of the throws—she can succeed on exactly 3 throws or on all 4. Find the total probability by determining the chance of each of these two distinct outcomes and adding them together.

Start with the chance of succeeding on all 4 throws. Since Leila’s chance of success is 1/5, multiply out her chance of succeeding 4 times in a row: 1/5 × 1/5 × 1/5 × 1/5 = 1/54

You can now eliminate answer A; it can’t be correct, because you haven’t yet added in the possibility of succeeding on exactly 3 throws.

If Leila succeeds on exactly 3 of the throws, then she must have 1 missed throw. The chance of missing a throw is 1 – 1/5 = 4/5, so multiply the individual probabilities to get the total probability: 1/5 × 1/5 × 1/5 × 4/5 = 4/54

However, this is the probability of only one specific outcome: hit, hit, hit, miss. Since she doesn’t need to make her throws in this specific order, you'll also need to consider the other possibilities: hit, hit, miss, hit hit, miss, hit, hit miss, hit, hit, hit

There are 4 different ways that Leila can get 3 hits and 1 miss. (Note that this wasn’t an issue when calculating the chance of succeeding on all 4 throws, because there is only one way to do that.) Therefore, her chance of successfully making exactly 3 out of 4 throws is 4(4/54)=16/54.

Finally, add the two values together to find the probability that Leila succeeds on at least 3 of the throws:

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25 Aug 2019, 02:19

Archit143 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

Given: Leila is playing a carnival game in which she is given 4 chances to throw a ball through a hoop.

Asked: 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?

Probability (chances) that she will succeed on at least 3 of the throws = \(^4C_3 * (1/5)^3(4/5) + (1/5)^4 = 4 * 4/5^4 + 1/5^4 = 17/5^4\)

Re: Leila is playing a carnival game in which she is given 4
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26 Aug 2019, 03:10

Archit143 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

success; 1/5 and fail ; 4/5 so for 4 chances we have ; case 1 all in ; (1/5)^4 case 2 ; 3 in and 1 out ; 4c3* ( 1/5)^3*(4/5) case 1+ case 2 ; 17/5^4 IMO E

gmatclubot

Re: Leila is playing a carnival game in which she is given 4
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26 Aug 2019, 03:10