Leila is playing a carnival game in which she is given 4

Question

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

A. 1/5^4
B. 1/5^3
C. 6/5^4
D. 13/5^4
E. 17/5^4

Bunuel · Accepted Answer

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 \(P=C^3_4*(\frac{1}{5})^3*\frac{4}{5}+(\frac{1}{5})^4=\frac{17}{5^4}\).

Answer: E.

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Hope it helps.

BrentGMATPrepNow · Answer

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
= E

BrushMyQuant · Answer

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 of Success =  \frac{1}{5} 
Probability of Failure= 1 - Probability of Success = 1-  \frac{1}{5}  =  \frac{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 at least 4 throws =
P(FSSS) + P(SFSS) + P(SSFS) + P(SSSF) + P(SSSS) =
 {(\frac{4}{5}) *(\frac{1}{5}) *(\frac{1}{5}) *(\frac{1}{5})} + {(\frac{1}{5}) *(\frac{4}{5}) *(\frac{1}{5}) *(\frac{1}{5})} + {(\frac{1}{5}) *(\frac{1}{5}) *(\frac{4}{5}) *(\frac{1}{5})} + {(\frac{1}{5}) *(\frac{1}{5}) *(\frac{1}{5}) *(\frac{4}{5})} +{(\frac{1}{5}) *(\frac{1}{5}) *(\frac{1}{5}) *(\frac{1}{5})}  =
  (\frac{1}{5})*(\frac{1}{5})*(\frac{1}{5})* { (\frac{4}{5}) + (\frac{4}{5}) + (\frac{4}{5}) + (\frac{4}{5}) + (\frac{1}{5}) } 
=  \frac{17 }{ 5^4}

So,  Answer will be E 
Hope it helps!

Archit143 · Answer

{4c3 * (1/5)^3*(4/5)^1}+{1/5*1/5*1/5*1/5}
16/5^4 +1/5^4
17/5^4

Am i correct

BrushMyQuant · Answer

Yes... i don't use the Combination formula much so i was thinking for a second.

IvyLeague56 · Answer

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

Phoenix72 · Answer

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?

Bunuel · Answer

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

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

Hope it's clear.

SaraLotfy · Answer

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.

ramannanda9 · Answer

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

sgrover18 · Answer

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

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

Regards,
Shradha

Poorvasha · Answer

P (that she succeeds) = 1/5
P(of success exactly 3 times) = 1/5*1/5*1/5*4/5 = 4/5^4. Now, these can occur in 4!/3!*1! = 4. hence, P(success 3 times)= 4*4/5^4
P (success all 4 times)= 1/5^4
Answer= 16/5^4+1/5^4 = 17/5^4 (E)

Archana24 · Answer

Is it possible to solve this using complement rule.

Posted from my mobile device

testcracker · Answer

at least 3 wins out of 4 chances can happen as follows 3 wins and 1 loss it can happen in 4 ways OR 4 wins it can happen in 1 way thus (1/5)^3 * (4/5) * 4 + (1/5)^4 * 1 17 _____ = the answer 5^4 thanks

dabaobao · Answer

Official Solution (Credit: Manhattan Prep) 
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:

1/15^4+16/15^4 = 17/15^4

The correct answer is E.

Kinshook · Answer

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

IMO E

Archit3110 · Answer

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

CEdward · Answer

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

A. 1/5^4
B. 1/5^3
C. 6/5^4
D. 13/5^4
E. 17/5^4

Let S be success and F be fail

Possibility one: 
SSSF

P(SSSF) = (1/5)^3 x 4/5 = 4/625...but this can happen 4!/3! = 4 ways so ... 4/625 x 4 = 16/625

Possiblity two: 
SSSS

P(SSSS) = (1/5)^4 = 1/625

16/625 + 1/625 = 17/625 = 17/5^4

E.

Ghoskrp3 · Answer

Successful in atleast 3 throws means, either successful in 3 or successful in 4 throws:

3 throws success =>  \frac{1}{5} *  \frac{1}{5} *  \frac{1}{5} *  \frac{4}{5}  multiplied by 4 (as there are only 4 ways in which this outcome would emerge)
=> (16)* 1/5^4

and 4 throws success =>  \frac{1}{5} *  \frac{1}{5} *  \frac{1}{5} *  \frac{1}{5} 
 => 1/5^4
 
 Hence, (16)*1/5^4 + 1/5^4 = (17)*1/5^4

Answer: E

Leila is playing a carnival game in which she is given 4

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Leila is playing a carnival game in which she is given 4

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