Mike recently won a contest in which he will have the opport

I didn't quite get this one

This is my procedure

OK if he shot one time only probability will be 'p'

If he makes at least two shots then probability will be: Probability of making all three shots + Probability of making 2 shots and missing 1 shot

Let's see probability of making all three is p^3

Probability of making 2 out of 3 is 3C2 * p^2* (1-p)

So question goes down to is p> 3(p^2)(1-p) + p^3?

Is 2p^2 - 3p + 1 <0?

Statement 1

If we replace 0.7 for p we get that yes, it is always negative hence true

Sufficient

Statement 2

Clearly insufficient.

Could someone explain if I'm wrong somewhere? I'm mainly concerned about the formulation of the question rather than in the solving part but still could have made an error

Many thanks
Cheers
J

First of all, my friend, I think you have made a mistake in opening the brackets. I got it as 3p^2- 2p^3 -p < 0 (I'm bad at algebra )
Secondly, Why to complicate this question? By reading it you can understand it will involve cubes and squares + fractions. So its rather better to put your own values (under the permissible limit). My friend, solve question strategically and not mathematically

Third, this may not sound the best way to solve but you can do a little thinking before diving in with values.
A probability of < 0.7 can mean it could 0.1 or 0.6
if its 0.1 ----> if you are very lucky you might get a throw in once chance but 2 out 3 throws could mean tough for you.
if its 0.6 ----> More than half of the time you get a shot. You might or might not get it in one chance but you are better off with taking three shots as you are aware half of the times you get the shot

INSUFFICIENT

p > 0.6
Obviously here you can easily say, your probability is quite good. Although you don't get every shot (p=1, read question he occasionally misses the shot so p =1 is not possible ) but you mostly get the shot so you better go for three shots than one.

It didn't take me much time to think this obviously reading/writing this might seems like a long process but it seriously doesn't even take more than half a minute. After this, you can assume some value to confirm


SUFFICIENT

Note that had they not written "occasionally he misses the shot" it would have been tough as p> 0.6 could mean p=1
So if you always get a shot than taking one or three doesn't matter.

Thanks Buddy, agree that algebraic approach to this question could prove painful?

Bunuel, do you want to take a stab at this one and show us how you roll?

Thanks
Cheers!
J

could you please explain how you got this? \(C^2_3p^2(1-p)+p^3=3p^3(1-p)+p^3=p^2(3-2p)\).

At least 2 hits in 3 shots means 2 or 3 hits:

The probability of 2 hits in 3 shots is \(P(HHN) = C^2_3p^2(1-p)\), we are multiplying by \(C^2_3=3\) because HHN scenario can occur in 3 ways: HHN, HNH, NHH.

The probability of 3 hits in 3 shots is \(P(HHH) = p^3\).

Thus the overall probability for at least 2 hits in 3 shots is \(C^2_3p^2(1-p)+p^3\).

Hope it's clear.

Hi Bunuel,

why do we need to consider the case HHN , i mean once he gets two hits ......wouldnt that suffice??

P ( winnning in 3 attempts) :
P ( hitting target in all 3 attempts) + P (hitting target in atleast 2)

= p^3 + 3 (p^2 (1- p ))
= p^3 + 3p^2 - 3p^3
= -2p^3 + 3p^2

Rephrase question is p> -2p^3 + 3p^2
after simpifying we get :

is 2p^2 - 3p + p > 0 ?
boils down to is p> 1/2?

Hence B

Yes, you're right here. We get the same answer either way. Considering just the option 'HH' includes both 'HHM' and 'HHH'; Bunuel just counted them separately in his answer.

Notice that the probability of HHM + probability of HHH = probability of HH:
\(p^2(1-p) + p^3 = p^2 - p^3 + p^3 = p^2.\)
Careful though! If you just consider HH and NOT HHM, then DO NOT include HHH, as it's already included in HH.

So yes, you're correct; it's just a matter of which approach you like best.

Can you please clarify why order does matter here??I solved this problem after solving a problem similar to it,the one where triplets Adam,Bruce and Charlie enter a triathlon.There we needed to find probability that atleast 2 triplets will win a medal.Here we have to find probability of at least two throws going correctly. But still in this problem we are considering order,but not in the triplets problem. Please clarify.

Forget at least 2. We are talking now about 2 hits in 3 throws: HHN. We can have 2 hits in 3 throws in following ways: HHN, HNH, NHH. Each of which has its probability. Sorry, cannot explain any better.

Awesome. Good to know that my method of solving it matched with yours.

Do we have some more problems in which we can apply same binomial distribution concept?

Thanks,
Gaurav

Some questions to practice:
what-is-the-probability-that-a-field-gun-will-hit-thrice-on-127334.html
there-is-a-90-chance-that-a-registered-voter-in-burghtown-56812.html
combination-ps-55071.html
the-probability-that-a-family-with-6-children-has-exactly-88945.html?

Hope it helps.

One thing to clarify, 3 free shots in the question means - one shot after another shot and then one more shot, am I right?

__________
Yes. How else?

Hi Bunuel,

Firstly, thank you for all the work you have done on this site. my doubt is why are we calculating all three wins i.e p^3; shouldn't we just write p^2 for two consec wins as the third win wont matter?

Check here: mike-recently-won-a-contest-in-which-he-will-have-the-opport-86901.html#p1357313

The question requires us to determine whether Mike's odds of winning are better if he attempts 3 shots instead of 1. For that to be true, his odds of making 2 out of 3 must be better than his odds of making 1 out of 1. There are two ways for Mike to at least 2 shots: Either he hits 2 and misses 1, or he hits all 3:

In order for this inequality to be true, p must be greater than .5 but less than 1 (since this is the only way to ensure that the left side of the equation is negative). But we already know that p is less than 1 (since Mike occasionally misses some shots). Therefore, we need to know whether p is greater than .5. If it is, then the inequality will be true, which means that Mike will have a better chance of winning if he takes 3 shots.
Statement 1 tells us that p < .7. This does not help us to determine whether p > .5, so statement 1 is not sufficient.
Statement 2 tells us that p > .6. This means that p must be greater than .5. This is sufficient to answer the question.
The correct answer is B: Statement (2) alone is sufficient, but statement (1) alone is not sufficient.

Refer attachment to see the calculation.