Biomial Probability (1 Viewer)

[lyounamu]

A dental inspector finds that about 20% of children of a certain area have tooth decay. What is the probablity that out of 6 children examined, only the first, third and fifth will have tooth decay?

I got 0.2 x 0.8 x 0.2 x 0.8 x 0.2 x 0.8 = 64/15625. What's the problem with my working out? The answer is 64/3125.

I used my working out for the other question but I got it right.

 

[lyounamu]

This is another question. I have done part i), ii) and iii) but I encountered a problem for part iv)

Three uniform tetrahedrea (triangular pyramids) each has one face black, another white, another red and the other green. When tossed onto a table, three faces of ech tetrahedron can be seen. What is the probability that (part iv) 3 white faces and only 1 green face can be seen?

What I did here was that I found the P(3 white faces) x P(only 1 green face). The answer was remotely close. Help is appreciated.

 

[lyounamu]

Anyone?

 

[midifile]

Whats the answer to the second 1?

 

[lyounamu]

Whats the answer to the second 1?

3/32. Did you get it? Post up solution please if that's the case.

What about the first one? You reckon you got it?

 

[Iruka]

About the 2nd Q, the way I was thinking is this:

You can see all but one face of each tetrahedra (the one that is facing down) so if you can only see 1 green one, then clearly 2 of the tetrahedra must land with green facing down. Of the remaining tetrahedron, it must have green facing up, but it must also have white facing up, otherwise there wont be three white faces showing. Now, for a tetrahedron to land with green down, there is 1/4 chance. There is 2/4 chance that it will land with neither white nor green facing down, and three are three different ways to choose the tetrahedron that does not have green or white facing down.

So, P(E) = (1/4)^2 x 2/4 x 3 = 3/32

 

[lyounamu]

About the 2nd Q, the way I was thinking is this:

You can see all but one face of each tetrahedra (the one that is facing down) so if you can only see 1 green one, then clearly 2 of the tetrahedra must land with green facing down. Of the remaining tetrahedron, it must have green facing up, but it must also have white facing up, otherwise there wont be three white faces showing. Now, for a tetrahedron to land with green down, there is 1/4 chance. There is 2/4 chance that it will land with neither white nor green facing down, and three are three different ways to choose the tetrahedron that does not have green or white facing down.

So, P(E) = (1/4)^2 x 2/4 x 3 = 3/32

Wow, you are awesome! I cannot stop admiring your extreme mathematical intelligence. I get it now. I just kept thinking that the question was related to the earlier 3 questions where I solved it via binomial probabiltiy method. That question really makes you stop and think before you actually write stuff down.

Again, thanks a lot!

 

[midifile]

3/32. Did you get it? Post up solution please if that's the case.

What about the first one? You reckon you got it?

What I did was think about the side that was face down. The green face has to be face down for 2 tetrahedra (as there is only one green face up). The third one can have either the black or red face down.

P(3 white + 1 green) = 1/4 x 1/4 x 2/4 x 3!/2!1!
= 3/32

I'm not sure about the first one. It looks fine to me. Maybe there is something wrong with the answer or something we are both missing
.

 

[midifile]

About the 2nd Q, the way I was thinking is this:

You can see all but one face of each tetrahedra (the one that is facing down) so if you can only see 1 green one, then clearly 2 of the tetrahedra must land with green facing down. Of the remaining tetrahedron, it must have green facing up, but it must also have white facing up, otherwise there wont be three white faces showing. Now, for a tetrahedron to land with green down, there is 1/4 chance. There is 2/4 chance that it will land with neither white nor green facing down, and three are three different ways to choose the tetrahedron that does not have green or white facing down.

So, P(E) = (1/4)^2 x 2/4 x 3 = 3/32

Damn you got there before me

 

[lyounamu]

What I did was think about the side that was face down. The green face has to be face down for 2 tetrahedra (as there is only one green face up). The third one can have either the black or red face down.

P(3 white + 1 green) = 1/4 x 1/4 x 2/4 x 3!/2!1!
= 3/32

I'm not sure about the first one. It looks fine to me. Maybe there is something wrong with the answer or something we are both missing
.

Thanks for your answer. That's another awesome working out.

I agree with ya. I think my textbook probably had a wrong answer at the back. It occurs quite casually. Thanks.

 

[independantz]

A dental inspector finds that about 20% of children of a certain area have tooth decay. What is the probablity that out of 6 children examined, only the first, third and fifth will have tooth decay?

I got 0.2 x 0.8 x 0.2 x 0.8 x 0.2 x 0.8 = 64/15625. What's the problem with my working out? The answer is 64/3125.

I used my working out for the other question but I got it right.

You forgot to take into account the number of awys of choosing the kids.

6C3-(5C1x3C1x1C1)=5

5x64/15625=64/3125

 

[lolokay]

You forgot to take into account the number of awys of choosing the kids.

6C3-(5C1x3C1x1C1)=5

5x64/15625=64/3125

how are you supposed to choose the kids when it already gives you the order that they're in? can you explain this further? :/

 

[independantz]

Well probability is desirable/possible , so even though they tell you the order, you'd still have to take that into account for the "desirable" part.

 

[lolokay]

I'm still a bit lost as to why you can't just do the "0.2 x 0.8 x 0.2 x 0.8 x 0.2 x 0.8 = 64/15625", or what "6C3-(5C1x3C1x1C1)=5" means.

 

[lyounamu]

You forgot to take into account the number of awys of choosing the kids.

6C3-(5C1x3C1x1C1)=5

5x64/15625=64/3125

I think that makes sense to some extent. But I would ask my teacher just to be on a safe side. Thanks for the reply though.