Any time you see boxes in the answers in a quant problem, you should try to do the easy ones first, since there must be at least one box that's true. If they happen to all be not true, then the last, most difficult one must be true, and you didn't have to deal with it.

Let's look at A first. The probability of him rolling 4 strikes in a row is .4x.4x.4x.4 or .4^4. 4^4 should be on your list of numbers to memorize and it equals 256, so .4^4 equals .0256, which is less than 3%. So A is out.

OK now B. I googled up the original question and it's supposed to say "less than 10%". The odds of him missing 4 times in a row is similarly .6^4. Personally I don't think you need to memorize the powers of 6 past 6 cubed. What I did was I wrote down .36 x .36, which would of course be the same as .6^4. You almost never have to actually multiply numbers on a piece of paper and this time I didn't either. I noticed that .36 is a little more than a third. So a third of .36 should be .12 and thus the odds of him missing four times in a row should be a bit over 12%, so clearly B is out.

C looks obviously wrong if you haven't done stuff like this much, but I have seen cases in which something counter-intuitive like this is actually true. I knew that to verify it I'd need to figure out the odds of him getting one strike and the odds of him getting two, but that seemed time-consuming so I skipped it for now.

D's turn. If, in probability or combinatorics, you see any phrasing like "at least one" or "at most 3", in general you will want to find the opposite of what they asked and then subtract that from one to find what they asked. In this case they said "2 or more." I could find the odds of him getting 2 strikes, and 3 strikes, and 4 strikes, and then add them. Or, I could just find the odds of 0 strikes and 1 strike and then add them, and we already found the odds of 0 strikes. So let's find the odds of him getting one strike. Assuming he got it on the first try and then missed the next 3, we can say the odds would be .4x.6x.6x.6. .6 cubed is .216, multiplied by .4 gets us .0864. But we have to remember that he could've gotten his strike on the 2nd, or 3rd, or 4th try. Since there are 4 scenarios we should multiply .0864 by 4. I don't want to do that. Let's round it to .09 and multiply by 4, getting us .36, which is an overestimate of the odds. That's important. Then we can add the odds of him missing every time, which we'd figured out earlier was a bit over .12. If we round that up to .13, again an overestimate, then when we add them we get .49, which we know is an overestimate of the odds that he got 0 or 1 strike. What does this mean? That the odds he got 2 or more must be 1 - .49, which is .51, or over half. And we know it's even greater than that. So D is out.

Hey guess what? It's C and we didn't even have to do it.

Tom is equally likely to roll exactly 1 strike as to roll exactly 2 strikes in those 4 frames.

the probability of Tom rolling exactly 1 strike, is a strike in any one of the rolling session
{4/10, 1, 1, 1} + {1, 4/10, 1, 1} + {1, 1, 4/10, 1} + {1, 1 1, 4/10} = 4 * (4/10) = 16/10

the probability of Tom rolling exactly 2 strike, is a strike in any two of the rolling session
{4/10, 4/10, 1, 1} + {1, 4/10, 4/10, 1} + {1, 1, 4/10, 4/10} + {4/10, 1 1, 4/10} = 4 * (4/10) = 16/10

C is the answer

You got the C as the answer right but the solution is not.

the probability of Tom rolling exactly 1 strike, is a strike in any one of the rolling session is NOT:
{4/10, 1, 1, 1} + {1, 4/10, 1, 1} + {1, 1, 4/10, 1} + {1, 1 1, 4/10} = 4 * (4/10) = 16/10

it is:
{4/10, 0.6, 0.6, 0.6} + {0.6, 4/10, 0.6, 0.6} + {0.6, 0.6, 4/10, 0.6} + {0.6, 0.6, 0.6, 4/10} = 4 * (4/10) * (0.6)^3 = 4*0.0864

AND

the probability of Tom rolling exactly 2 strike, is a strike in any two of the rolling session is NOT:
{4/10, 4/10, 1, 1} + {1, 4/10, 4/10, 1} + {1, 1, 4/10, 4/10} + {4/10, 1 1, 4/10} = 4 * (4/10) = 16/10

it is:
{0.4, 0.4, 0.6, 0.6} + {0.4, 0.6, 0.4, 0.6} + {0.4, 0.6, 0.6, 0.4} + {0.6, 0.4, 0.4, 0.6} + {0.6, 0.4, 0.6, 0.4} + {0.6, 0.6, 0.4, 0.4} = 6 * (0.4)^2 * (0.6)^2


Now, 4 * 0.4 * 0.6^3 = 6 * 0.4^2 * 0.6^2

Hence, C is correct.

For option D:

probability of getting two or more strikes out of 4 frames is same as 1 - (probability of getting no strike or getting exactly one strike)

1 - [(0.6^4) + 4*(o.4)*(0.6)^3]

So, you think when Tom Strike on the first sessions he shouldn't strike on the following three {4/10, 0.6, 0.6, 0.6}? Why not count the combinations that Tom strick on the first and the following sessions could be anything? is the word "exactly" exclude these combinations? I still think that Tom needs only to strike in one session the following three could be anything.. hence {4/10, 1, 1, 1} is correct.

As it is mentioned that Tom has to strike ‘exactly’ once, we have to take the probability of 0.6 for all other frames. If the question had mentioned ‘at least’ once, then yours would have been correct.

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I don't understand the question! Could you please tell me what he meant by this question in simple language?

Is there a reason why 2 strikes has the same probability as a single strike? It looks wrong but by calculations it is indeed equal. Is there a normal distribution of probabilities trend going on? Can we extrapolate the probability for 3 strikes from this trend?

It certainly seems strange to think those two probabilities could be the same.
However, there are similar examples to be found.

For example, if we flip a "fair" (i.e., the probability of getting heads = the probability of getting tails = 1/2) coin 4 times, the probability of getting exactly 1 head is the SAME has the probability of getting 3 heads.
Now this phenomenon might not be that surprising, since getting 3 heads and 1 tail is just as likely as getting 1 head and 3 tails.

However, when we consider the fact that, in the case of bowling, P(getting a strike) does NOT equal P(not getting a strike), the results of the bowling probabilities aren't as surprising.

A simpler explanation:

The probability of Tom rolling exactly 1 strike in 4 frames is 0.4 x 0.6 x 0.6 x 0.6 x 4C1 = 0.3456
The probability of Tom rolling exactly 2 strikes in 4 frames is 0.4 x 0.4 x 0.6 x 0.6 x 4C2 = 0.3456
Since both the above results are the same, we can say that C is true.

could any one explain this statement for me (Tom will roll 2 or more strikes less than half the time.)

It means that we have to find the probability when Tom rolls at least 2 strikes in a rolling session.

Why do you use combinations here rather than permutations? To me, it seems like order does matter, though I know combinations are actually right here. Like if I draw out the possibilities of rolling exactly 2 strikes, it's

X X_ _ , _ _ X X , _ X X _ , X _ _ X , X _ X _ , _ X _ X

And to me this looks like an order matters scenario.

I found the quickest way to verify the statements was to model the number of strikes as a binomial random variable with (n = 4) and (p = 0.4).

* (P(\text{4 strikes}) = (0.4)^4 = 0.0256), so statement 1 is false.
* (P(\text{0 strikes}) = (0.6)^4 = 0.1296), so statement 2 is also false.
* (P(\text{exactly 1 strike}) = \binom{4}{1}(0.4)(0.6)^3 = 0.3456).
* (P(\text{exactly 2 strikes}) = \binom{4}{2}(0.4)^2(0.6)^2 = 0.3456).

The equality in the last two probabilities is a neat coincidence that comes from the combination counts and the values of (p) and (1-p). So statement 3 is true.

Finally,
(P(\text{2 or more strikes}) = 1 - P(0) - P(1) = 1 - 0.1296 - 0.3456 = 0.5248),
which is greater than 50%, making statement 4 false.

So the correct answer is **C only**.