Theresa is a basketball player practicing her free throws. On her fir

There is a lot of computation to do in just 2 minutes. Is there a faster way to solve this problem? Thanks!

zy31

Hi,
How u got the probability of missing the basket?
Please explain..
thank you.

yeah this is a lot of computation for 2 minutes, in the actual test I would realize it is a large number so I would either take an estimated guess and go for D or E. Faster ways to solve it?

Good Question +1 for mikemcgarry
But the calculations are too many provided the time pressure. I got the correct answer, but is this to be expected on the GMAT?

Dear ShashankDave,

Thank you for the kudos.

Here's what I'll say. First of all, I would say a question of this sort would be, possibly, at the way outer limit of what the GMAT might ask. In other words, if you are acing everything on the quant sections, and the CAT is throwing the hardest questions at you to see whether there's anything you might get wrong, perhaps you would see something like this. Even then, that's a maybe.

Also, on the time pressure, I say, there's a certain issue of level of mathematical fluency. If you have to pause and think out each step laboriously, then the calculation would take a long time. If you see patterns and are able to make connections quickly, it would be possible to do this in under 2 mins, but only a tiny percent of the population has that kind of mathematical facility. For example, all the fractions have 5 in the denominator, and all the answers have \(5^5\) in the denominator, so really, we can ignore denominators, and just look at numerators. If you have to write out all the patterns in words or letters first (e.g. Y, Y, N, Y, Y) and then write numbers, that would take longer for simply writing the numbers directly and recognizing what pattern of made vs. missed baskets these numbers represent. Also, there are all sorts of number sense tricks for finding the products and the sum. So, the better you are at making lightning-fast connections and spotting patterns, the quicker this and many other hard questions will be. I don't know whether this answer is helpful.

Let me know if you have any other questions.

Mike

Dear manojhanagandi,

I see nobody replied, so I'm happy to help.

In this blog,
GMAT Math: the Probability “At Least” Question
read the section about the "complement rule." That's one of the most important probability rules to understand.

Let me know if you have any questions.
Mike

This question involves lots of calculation, which requires carefulness but is feasible in "4-5 mins".

Step 1: To reduce the chance of making calculation mistakes, I draw a table of probability before jumping into the calculation. (this is probably the most important step in solving the question)

First throw: Succeed 60% = 3/5 - Fail 40% = 2/5
After 1st throw:
If previous throw:
a. Succeed, then this current throw has chance: Succeed 80% = 4/5 - Fail 20% = 1/5
b. Fail, then this current throw has chance: Succeed 40% = 2/5 - Fail 60% = 3/5
* I convert everything to have 5 as denominator as the options provide hints of having 5^5 in denominator (converting saving time to reducing everything from 10 to 5)

Step 2: Calculation:

There are 5 ways the person has at least 4 successful free throws: 4 ways of failing ONLY 1 at any of the 5 throws AND 1 way of succeeding in all 5 throws.
Since each throw is influenced by previous throw result and influence next throw result, calculating 1 way and multiplying by 4 is a WRONG way to calculate the probability of having exactly 4 successful throws.

SSSSF: Prob = (3/5)*[(4/5)^3]*(1/5)

SSSFS: Prob = (3/5)*[(4/5)^2]*(1/5)*(2/5)

SSFSS: Prob = (3/5)*(4/5)*(1/5)*(2/5)*(4/5)

SFSSS: Prob = (3/5)*(1/5)*(2/5)*[(4/5)^2]

FSSSS: Prob = (2/5)*(2/5)*[(4/5)^3]

SSSSS: Prob = (3/5)*[(4/5)^4]

Since all options have 5^5 as dominators and my calculation is on the track, I calculate only the numerator.
Numerator = 3*(4^3) + 3* (4^2)*2 + 3*(4^2)*2 + 3*2*(4^2) + (2^2)*(4^3) + 3*(4^4)
= (4^2)*(12+6+6+6+16+48)
= 16* 94 = 1504

Long calculation, but fulfilling when the answer is correct. During the exam, I may take max 3-4 mins to calculate and probably skip if my result doesn't match any of the options.

--
Please give a kudos if you find this post useful.

wouldnt it be easier to calculate the change of missing all five and missing 4 out of five then subtracting that from one??

Then you're gonna get a number bigger than the correct answer, because your result will include 2 redundant cases - miss 3/5 and miss 2/5, while your answer should be comprised of miss 1/5 and miss 0/5 only!

After all, your suggested solution may sound more lengthy than the one suggested by Mike.

too much calculation!!!.. does such questions really come in GMAT....

Dear Ruchita1907,

I'm happy to respond.

Admittedly, this question would at the outer edge of what the GMAT might ask: if someone were acing everything else on the Quant section, and the CAT basically was throwing the kitchen sink at them, then it might throw a problem this details.

You see, it's a matter of mathematical fluency. If you have to think through each and every step, then this is a long and laborious problem. If you get to the point where you can see the patterns and race through them, then this question easily could be completed in under 90 seconds.

My friend, when you encounter something on the GMAT that seems hard--and later, when you encounter something in the business world that seems hard--don't back away and look for excuses to avoid it. Instead, ask yourself what you would have to do to rise to the challenge of it. Believe in your own capacity for success and follow that fiercely. Don't try to drag to the world down to what makes you comfortable: instead, push yourself past your own expectations to bring forth your own greatness.

Does all this make sense?
Mike

Pretty interesting problem involving calculations and deep thinking.

Two cases are possible here. 1) Atleast 4 throws with 1 not thrown chance.
2) All 5 chances are thrown into the basket.

Case 1) Now the single miss can be selected in 5 ways out of total 5 chances. It can be in first chance or second or third or fourth or fifth. Lets evaluate all.

First chance missed: \(\frac{2}{5}*\frac{2}{5}*\frac{4}{5}*\frac{4}{5}*\frac{4}{5} = \frac{4^4}{5^5}\)

Second chance missed: \(\frac{3}{5}*\frac{1}{5}*\frac{2}{5}*\frac{4}{5}*\frac{4}{5} = \frac{6*4^2}{5^5}\)

Third Chance missed: \(\frac{3}{5}*\frac{4}{5}*\frac{1}{5}*\frac{2}{5}*\frac{4}{5} = \frac{6*4^2}{5^5}\)

Fourth Chance missed: \(\frac{3}{5}*\frac{4}{5}*\frac{4}{5}*\frac{1}{5}*\frac{2}{5} = \frac{6*4^2}{5^5}\)

Fifth chance missed: \(\frac{3}{5}*\frac{4}{5}*\frac{4}{5}*\frac{4}{5}*\frac{1}{5} = \frac{3*4^3}{5^5}\)

Case 2) All chances were successful.

\(\frac{3}{5}*\frac{4}{5}*\frac{4}{5}*\frac{4}{5}*\frac{4}{5} = \frac{3*4^4}{5^5}\)

Total = \(\frac{4^4 + 3(6*4^2) + 3*4^3 + 3*4^4}{5^5}\)
= \(\frac{1504}{5^5}\)

OPTION: E

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