Probability Question

[rte.sushil]

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

Answer : E
Source Mahattan test series


Question is not very tough but still i missed an imp. point to consider while solving it. I didn't consider the order of her failure of her throw. I considered only once not four times as failure could happen at first place, 2nd place , 3rd or 4th.
i solved as 1/5^3 * 4/5 + 1/5^4 = 1/5^3

Somehow i am not able to apply the concept of order, so i missed it while solving. If the question were "take out 3 balls ,,,,5 green & 2 black, what is the probability that 3 balls are green " then i will simply do 3/7 , i will not consider order of balls taken out..


I think i am missing something critical? Any suggestions?

Thanks!!

[jlucero]

The big difference between your two problems is that in the test problem, we say "at least" something must happen. If I took 4 shots, there are 5 different things that could happen: I could make 0, 1, 2, 3, or 4 of those shots. But if you flipped a coin 4 times, you wouldn't say that flipping 4 heads is as likely as flipping 2 heads/2 tails. This is because there are more ways to flip 2H2T, than 4H. (it's the same reason you wouldn't say you have a 50% chance of winning the lottery- I could win or I could not win... there's a LOT more ways to not win the lottery, which is why the odds of winning are so low).

In your question, you need to have a very specific thing happen- you need to get a green ball, then another, than another. There's no other way to get three green balls. But if the question was picking 2 green and 1 black, you could get the black on the 1st, 2nd, or 3rd pick, which means there are 3 different ways to have that occur. Same thing with Leila, she can miss the 1st, 2nd, 3rd, or 4th shot OR make all 4 shots. Therefore we need to add the probabilities of all 5 of these outcomes:
4/5*(1/5)^3 +
4/5*(1/5)^3 +
4/5*(1/5)^3 +
4/5*(1/5)^3 +
(1/5)^4
=
16/5^4 + 1/5^4 = 17/5^4

[nocheivyirene]

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

P(S,S,S,...) = P(S,S,S,F) + P(S,S,S,S)

First, we calculate P(S,S,S,F). We should take note that the order of S,S,S,F could interchange to S,F,S,S or S,S,F,S, and etc. So, we need to permute 4 elements and divide it by 3 since we have 3 S's.

P(S,S,S,F) = (1/5)(1/5)(1/5)(4/5)(4!/3!1!) = (4)(4/5)(1/5)^3

Second, we calculate P(S,S,S,S) which will have just 1 arrangement all the time. P(S,S,S,S) = (1/5)^4

P(S,S,S,...) = (1/5)^3 [(4*4/5) + (1/5)] = 17/(5^4)

Answer: E

[tim]

you need to divide by 3! instead of 3 of course, but your calculations reflect that. nice job..

[nocheivyirene]

you need to divide by 3! instead of 3 of course, but your calculations reflect that. nice job..

Oh! Yes, thanks tim for pointing that out.

[tim]

no problem..

[AlexaD160]

How do you know when order matters in a probability problem? This wasn't obvious to me when seeing the question for the first time.

[RonPurewal]

when probabilities are multiplied together, order is ALWAYS taken into account.

so, if you are going to calculate anything by multiplying probabilities, you ALWAYS have to take into account the different possible orders in which things can happen.