- Thread starter
- Admin
- #1

- Jan 26, 2012

- 4,093

--------------------

Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter Jameson
- Start date

- Status
- Not open for further replies.

- Thread starter
- Admin
- #1

- Jan 26, 2012

- 4,093

--------------------

Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter
- Admin
- #2

- Jan 26, 2012

- 4,093

1) anemone

2) MarkFL

Solution (from anemone):

$\small=\text{P(5 Heads)}+\text{P(4 Heads 0 Tail 1 Edge)}+\text{P(4 Heads 1 Tail 0 Edge)}+\text{P(3 Heads 0 Tail 2 Edges)}+\text{P(3 Heads 2 Tails 0 Edges)}+\text{P(3 Heads 1 Tail 1 Edge)}$

$\;\;\small+\text{P(2 Heads 0 Tail 3 Edges)}+\text{P(2 Heads 1 Tail 2 Edge)}+\text{P(1 Heads 0 Tail 4 Edge)}$

$=0.5^5+\dfrac{0.5^4\cdot0.1\cdot5!}{4!}+ \dfrac{0.5^4\cdot0.4\cdot5!}{4!}+\dfrac{0.5^3\cdot0.1^2\cdot5!}{3!\cdot2!}+ \dfrac{0.5^3\cdot0.4^2\cdot5!}{3!\cdot2!}+\dfrac{0.5^3\cdot0.1\cdot0.4\cdot5!}{3!}+\dfrac{0.5^2 \cdot 0.1^3\cdot5!}{2!\cdot3!}+\dfrac{0.5^2\cdot0.4\cdot0.1^2\cdot5!}{2!\cdot2!}+\dfrac{0.5^1\cdot0.1^4 \cdot 5!}{4!}$

$=0.03125+0.03125+0.125+0.0125+0.2+0.1+0.0025+0.03+0.00025$

$=0.53275$

- Status
- Not open for further replies.