# Math Help - Basic probablity question.

1. ## 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)

2. Hello, puzzledwithpolynomials!

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.

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

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

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