- #1

mr_coffee

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A coin is tossed ten times. In each case the outcome H (for heads) or T (for tails) is recorded. (One possible outcome of the ten tossings is denoted THHTTTHTTH.)

I got a, d right i believe.

but I need someone to check if i did the others ones right.a. what is the total number of possible outcomes of the coin-tossing experiemnt?

I said: 2^10 = 1,024

**b. In how many of the possible outcomes are exactly 5 heads obtained?**

[# of outcomes with exactly 5 heads] = [total # of outcomes] - (10 choose 5)

This one I'm not sure what I'm suppose to do, when i say 10 choose 5, i mean you have a set of 10, and I'm wanting to choose 5 elements out of that set that are H but I'm not sure how to represeent this...

c. In how many of the possible outcomes is at least 8 heads obtained?

For this one would i do:

[# of outcomes with at least 8 heads obtained] = [total number of outcomes] - [number of outcomes that contain 8 heads]

= 1024 - (10 choose 8)

= 1024 - 10!/[8!(2!)]

= 1024 - 45 = 979

[# of outcomes with exactly 5 heads] = [total # of outcomes] - (10 choose 5)

This one I'm not sure what I'm suppose to do, when i say 10 choose 5, i mean you have a set of 10, and I'm wanting to choose 5 elements out of that set that are H but I'm not sure how to represeent this...

c. In how many of the possible outcomes is at least 8 heads obtained?

For this one would i do:

[# of outcomes with at least 8 heads obtained] = [total number of outcomes] - [number of outcomes that contain 8 heads]

= 1024 - (10 choose 8)

= 1024 - 10!/[8!(2!)]

= 1024 - 45 = 979

d. In how many of the possible outcomes is at least one head obtained?

[# of outcomes w/ at least 1 head] = [total # of outcomes] - [# of outcomes with no heads] = 1024 - 1 = 1023.

**e. In how many of the possible outcomes is at most one head obtained?**

I think i got this one, but the # of outcomes with 1 head i might have got wrong.

[# of outcomes with at most 1 head] = [# of outcomes with no heads] + [# outcomes with 1 head]

= 1 + (10 choose 1)

= 1 + 10!/[1!(10-1)!]

= 1 + 10!/9! = 11Any help would be great, I'm having issues figuring out how to choose once i figure out the method it seems.

I think i got this one, but the # of outcomes with 1 head i might have got wrong.

[# of outcomes with at most 1 head] = [# of outcomes with no heads] + [# outcomes with 1 head]

= 1 + (10 choose 1)

= 1 + 10!/[1!(10-1)!]

= 1 + 10!/9! = 11