Math questions

[jessica22]

If an ordinary coin is tossed six times, what is the probability of either all heads or all tails.

A. 1/4
B. 1/32
C. 1/3
D. 1/8
E. 1/16

Since this is independent event, I did this... 1/2 X 1/2 X 1/2 X 1/2 X 1/2 X 1/2 = 1/64...

But it's not one of the answer...plase helpppppp

#2 There are three consecutive even integers such that the third of the three even integers is 10 more than the sum of the first and the second. What is the third integers?

A. -8
B. -6
C. -4
D. -10
E. -2

let x=even integer, then the 3 integers will be x, x +2 , x +4

(x+4) +10 = (x+2) + x

x +14 = 2x +2
12=x

then the third integer has to be 14?????

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[dlink]

If an ordinary coin is tossed six times, what is the probability of either all heads or all tails.

A. 1/4
B. 1/32
C. 1/3
D. 1/8
E. 1/16

Since this is independent event, I did this... 1/2 X 1/2 X 1/2 X 1/2 X 1/2 X 1/2 = 1/64...

But it's not one of the answer...plase helpppppp

#2 There are three consecutive even integers such that the third of the three even integers is 10 more than the sum of the first and the second. What is the third integers?

A. -8
B. -6
C. -4
D. -10
E. -2

let x=even integer, then the 3 integers will be x, x +2 , x +4

(x+4) +10 = (x+2) + x

x +14 = 2x +2
12=x

then the third integer has to be 14?????

Answer for # 1 is 1/64.

Solution for # 2:

let x=even integer, then the 3 integers will be x, x +2 , x +4

(x+4) = (x+2) + x + 10

x +4 = 2x + 12
-8 = x

The third integer is x + 4 = -4

Answer is C.

 

[FebDAT]

j

 

[BKN]

If an ordinary coin is tossed six times, what is the probability of either all heads or all tails.

A. 1/4
B. 1/32
C. 1/3
D. 1/8
E. 1/16

Since this is independent event, I did this... 1/2 X 1/2 X 1/2 X 1/2 X 1/2 X 1/2 = 1/64...

But it's not one of the answer...plase helpppppp

Since the question said either all heads or all tails the first toss constrains the rest (in other words the first toss can be either, the rest must agree with it).

So 1*1/2*1/2*1/2*1/2*1/2 = 1/32

 

[scribbles]

Since the question said either all heads or all tails the first toss constrains the rest (in other words the first toss can be either, the rest must agree with it).

So 1*1/2*1/2*1/2*1/2*1/2 = 1/32

Hey BKN, I was having trouble understanding that prob too.

can you explain about constraining?

 

[fancymylotus]

Hey BKN, I was having trouble understanding that prob too.

can you explain about constraining?

The problem says "the probability of tossing *ALL* heads or *ALL* tails" This means that depending on what the first coin toss gives, the rest have to match that. Therefore, if the first toss gives a head, the rest will have to be heads. If the first gives tails, the rest will have to be tails. So, 1(represents first coin)*1/2(second coin toss-50% chance of heads or tails but you only want one of them depending on what you got the first time)*1/2, etc etc.

 

[jessica22]

thanks u guys... that was helpful. thank you.

 

[scribbles]

The problem says "the probability of tossing *ALL* heads or *ALL* tails" This means that depending on what the first coin toss gives, the rest have to match that. Therefore, if the first toss gives a head, the rest will have to be heads. If the first gives tails, the rest will have to be tails. So, 1(represents first coin)*1/2(second coin toss-50% chance of heads or tails but you only want one of them depending on what you got the first time)*1/2, etc etc.

Ahhh, now i see. And all this time I thought the book had it wrong. Thanks Amsie and thanks Jessica for asking the questions

 

[zacky2005]

For problem 1, you can also use the following approach:

The probabilty of tossing 6 heads in a row, P1 = 1/64

The probabilty of tossing 6 tails in a row, P2 = 1/64

P = P1 + P2 = 1/32

 

[BKN]

Hey BKN, I was having trouble understanding that prob too.

can you explain about constraining?

Amsie did a nice job of explaining what I meant in the problem. Sorry if I wasn't clear.

The term constraint in statistics refers to how many values of the parameters (variables) are known in a function before the equation is solved (that is the other variables are constrained to be a certain value). That is the "degrees of freedom", another unfortunate piece of statistical jargon.

So in the question we are discussing, if we are to meet the conditions set the first toss constrains all the other tosses to be the same.
Zacky's approach which is just to count all possible combinations works too, but it becomes difficult for using complex problems. You'd have to list them all. For example, if anyone wants to tackle it:

What is the probability of tossing a coin six times and getting 5 of one result and 1 of the other?

 

[ecrown]

Amsie did a nice job of explaining what I meant in the problem. Sorry if I wasn't clear.

The term constraint in statistics refers to how many values of the parameters (variables) are known in a function before the equation is solved (that is the other variables are constrained to be a certain value). That is the "degrees of freedom", another unfortunate piece of statistical jargon.

So in the question we are discussing, if we are to meet the conditions set the first toss constrains all the other tosses to be the same.
Zacky's approach which is just to count all possible combinations works too, but it becomes difficult for using complex problems. You'd have to list them all. For example, if anyone wants to tackle it:

What is the probability of tossing a coin six times and getting 5 of one result and 1 of the other?

Is the answer 3/32

 

[BKN]

Is the answer 3/32

Using counting method: Total number of possibilities 2^6 = 64 then:

HTTTTT
THTTTT
TTHTTT
TTTHTT
TTTTHT
TTTTTH
THHHHH
HTHHHH
HHTHHH
HHHTHH
HHHHTH
HHHHHT = 12/64 = 3/16. Tomorrow, if I'm feeling ambitious I'll show you how to do this with a formula that makes it much easier.

 

[seoul]

You may actually use the permutation formula to get it solved:

For 5 heads and 1 tail in a row, there're 5 heads that look alike and one tail that stand alone, leading to the possible arrangement of:

6!/5!1! = 6 ---- (1)

The same shall apply to 5 tails and 1 head in a row:

6!/5!1! = 6 ---- (2)

P = (6 + 6)/64 = 3/16

ecrown's answer would be correct if the question was referring to one of the two incidents above.

 

[howui3]

You may actually use the permutation formula to get it solved:

For 5 heads and 1 tail in a row, there're 5 heads that look alike and one tail that stand alone, leading to the possible arrangement of:

6!/5!1! = 6 ---- (1)

The same shall apply to 5 tails and 1 head in a row:

6!/5!1! = 6 ---- (2)

P = (6 + 6)/64 = 3/16

ecrown's answer would be correct if the question was referring to one of the two incidents above.

in case people want to know the full formula:

nCk=n!/k!(n-k)!

where n is the total pool or the total of events (here its 6 toses)
and k is how many event you want (here its 5 tails or 5 heads)

plug in: 6C5=6!/5!(6-5)! = 6!/5!1! = 6 (here recall that you can have 5 tails or 5 heads so multiply by 2)

hope this helps.

check out this post for more probability questions:

http://forums.studentdoctor.net/showthread.php?t=269878

good luck.

 

[BKN]

in case people want to know the full formula:

nCk=n!/k!(n-k)!

good luck.

Great. I was gonna take it one step further. Since most events, unlike a coin toss don't have a probability of .5, you can expand the binomial (things with two outcomes) using the permutation formula like this:

n = number of events
k= number of successes (outcome of interest)
n-k = number of failures
p = probability of a success in a single trial
q = 1-p = probability of failure in a single trial

so, the probablity of k successes in n trials if the probability of success in a single trial is p = (n!/k!n-k!)(p^k)(q^n-k)

Seems esoteric, but very useful in computing your chances of winning the lottery if you play once a week in a pool of 10 million entries. Expect to win the big prize something like once every 450 centuries.

I've used ^ to mean raised to power since the formatting doesn't seem to support superscripts.