# Basic probablity question.

• May 11th 2009, 12:06 PM
puzzledwithpolynomials
Basic probablity question.
Three coins are tossed. Find the probabilikty that they all land heads up for each known condition.

P(X|Y)=P(X)*P(Y)/P(Y)

6. The first coin shows a head.
My work: 2*2*2=8 possibilities
8 Pr 2=56 ways to choose that so it fits that one shows a head..
(1/56)*(1/8)/(1/8)=1/56?
7. At least one coin shows a head.
8. At least two coins show heads
...that's all I know how to work out so far.
Also, if somebody could explain how this logically makes sense, I'd really appreciate it! (Something like a tangible example or whatnot would be great)
• May 11th 2009, 01:43 PM
Soroban
Hello, puzzledwithpolynomials!

Quote:

Three coins are tossed.
Find the probabilikty that they all land Heads up for each known condition.

With only three coins, we can list the eight possible outcomes.

. . $\begin{array}{cccc} 1&H&H&H \\ 2&H&H&T \\ 3&H&T&H \\ 4&H&T&T \\ 5&T&H&H \\ 6&T&H&T \\ 7&T&T&H \\ 8&T&T&T \end{array}$

and we don't need any fancy formulas.

Quote:

6. The first coin shows a Head.

The first is a Head; there are 4 cases: . $HHH, HHT, HTH, HTT$

Among them, one case has all Heads.

Therefore: . $P(\text{all H }|\text{ 1st H}) \:=\:\frac{1}{4}$

Quote:

7. At least one coin shows a Head.

"At least one Head", there are 7 cases: . $HHH,HHT,HTH, HTT, THH, THT, TTH$

Among them, one case has all Heads.

Therefore: . $P(\text{all H }|\text{ at least 1 H}) \:=:\frac{1}{7}$

Quote:

8. At least two coins show Heads

"At least 2 Heads", there are 4 cases: . $HHH, HHT, HTH, THH$

Among them, one case has all Heads.

Therefore: . $P(\text{all H }|\text{ at least 2 H}) \:=\:\frac{1}{4}$