Probability

busprof

Friends lil help required with probablity.

Suppose i have a 80% Free throw percentage and i have 5 chances out of which i have to make 3 baskets.

I am trying to solve this by 2 different methods, getting correct answer from method 1 but not from second method.


Method 1.


5 nCr 3 = 10 Ways.

(Probability Of Success) * (Probability Of Failure) * Total ways

(.80 * .80 * .80) * (.20 * .20 ) * 10

= 20.480%



Method 2


Prob (a) = event : a true / total # of events

= 10 / 32
= 31.25%


-where total events are 2*2*2*2*2 = 32.



i dont know whats wrong with method 2 or maybe em missing something.

studyn

Ok, so in case 1, the way the formula works is that it first establishes that you made 3 throws and lost 2 throws. Hence (.80 * .80 * .80) * (.20 * .20 ).

Then it accounts the fact that you could have made the three throws in 10 different orders. So, it's (.80 * .80 * .80) * (.20 * .20 )*10.

What you have done in case 2 is equivalent to:
0.5*0.5*0.5*0.5*0.5*10.

In other words, case 2 assumes that your probability of making each throw is 50%, not 80% as stated in the question.

ALSO:

The case 1 method does not give you the probability of making at least 3 throws. It gives you the probability of making exactly 3 throws. I assume that you want this latter thing, since you're saying your answer was correct.

Jolyn

as others have pointed out, the first method is correct if you want to score exactly 3 baskets. If you want to score at least 3 baskets, you need to add probability of scoring 4 and 5. You will get 94.21%.

soddy1979

So do you guys mean that this question cannot be solved from 2nd method? if so pl tell me how.



by the way thanks everyone for your replies.

joemoran

ohai Wrote:
-------------------------------------------------------
> Huh what? I just explained what the second method
> is doing.

I absolutely agree with that explanation. Method 2 assumes that all outcomes have equal probabilities which is true when p = 1/2.