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The probability of Tom rolling a strike while bowling is 4
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22 Aug 2017, 02:20

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Question Stats:

The probability of Tom rolling a strike while bowling is 40% on any given frame. If Tom rolls 4 frames in a row, which of the following statements are true?

Indicate all such statements.

❑ The probability of Tom rolling a strike on all 4 frames is greater than 3%.

❑ The probability of Tom rolling no strikes in 4 frames is less than io %.

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

❑ Tom will roll 2 or more strikes less than half the time.

_________________

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Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
14 Jan 2018, 23:42

3

I could do it from process of elimination only

probability of tom rolling a strike is 0.4

probability of tom rolling a strike in all 4 frames = (0.4)^4 = 2.56% <3%

probability of tom rolling no strike is 1-0.4 = 0.6

probability of tom rolling no strike in all 4 frames = (0.6)^4 = 12.96% > 10%

probability of tom rolling 2 or more strike in 4 frames is

1-probability of tom rolling either 1 strike only or no strike in 4 frames

Scenarios for tom rolling 1 strike:

S,NS,NS,NS

NS,S,NS,NS

NS,NS,S,NS

NS,NS,NS,S

probability of 1st scenario is 0.4*0.6*0.6*0.6 = 0.0864

since all the 4 scenarios have exactly 3NS and 1S probability for each scenario = 0.0864

Therefore total probability for 1 strike in 4 frame = 3*0.0864=0.35

probability of tom getting no strike in 4 frames = 0.13

probability of tom getting no strike or 1 strike with 4 frames = 0.48

the probability for tom getting 2 or more than 2 strike with 4 frames: 1-0.48=0.52>0.5

(C)

_________________

probability of tom rolling a strike is 0.4

probability of tom rolling a strike in all 4 frames = (0.4)^4 = 2.56% <3%

probability of tom rolling no strike is 1-0.4 = 0.6

probability of tom rolling no strike in all 4 frames = (0.6)^4 = 12.96% > 10%

probability of tom rolling 2 or more strike in 4 frames is

1-probability of tom rolling either 1 strike only or no strike in 4 frames

Scenarios for tom rolling 1 strike:

S,NS,NS,NS

NS,S,NS,NS

NS,NS,S,NS

NS,NS,NS,S

probability of 1st scenario is 0.4*0.6*0.6*0.6 = 0.0864

since all the 4 scenarios have exactly 3NS and 1S probability for each scenario = 0.0864

Therefore total probability for 1 strike in 4 frame = 3*0.0864=0.35

probability of tom getting no strike in 4 frames = 0.13

probability of tom getting no strike or 1 strike with 4 frames = 0.48

the probability for tom getting 2 or more than 2 strike with 4 frames: 1-0.48=0.52>0.5

(C)

_________________

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Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
25 Jan 2018, 06:59

Could you also exclude D) by seeing that his chance of a strike is 40% for each frame.

Therefore he is more likely to hit less than two strikes after the second frame.

Thanks

Therefore he is more likely to hit less than two strikes after the second frame.

Thanks

Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
26 Jan 2018, 18:32

3

Expert Reply

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.

_________________

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.

_________________

Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
07 Apr 2018, 09:05

1

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

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

Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
09 Apr 2018, 23:28

1

HEcom wrote:

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

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.

Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
09 Apr 2018, 23:31

1

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]

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]

Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
11 Apr 2018, 03:47

novice07 wrote:

HEcom wrote:

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

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.

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.

Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
11 Apr 2018, 04:17

2

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.

Posted from my mobile device

Posted from my mobile device

Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
28 Jul 2018, 01:52

Carcass wrote:

The probability of Tom rolling a strike while bowling is 40% on any given frame. If Tom rolls 4 frames in a row, which of the following statements are true?

Indicate all such statements.

❑ The probability of Tom rolling a strike on all 4 frames is greater than 3%.

❑ The probability of Tom rolling no strikes in 4 frames is less than io %.

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

❑ Tom will roll 2 or more strikes less than half the time.

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

Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
05 May 2020, 04:11

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?

Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
05 May 2020, 16:26

1

Expert Reply

Zohair123 wrote:

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.

_________________

Re: The probability of Tom rolling a strike while bowling is 4
[#permalink]
01 Aug 2020, 23:42

1

novice07 wrote:

HEcom wrote:

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.

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.

gmatclubot

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