# Math Help - Help needed for a probablilty question

1. ## Help needed for a probablilty question

Ok so this is really bugging me, because I thought i finally understood it, but then my book is giving me different answers to what I'm getting.
Anyways, heres the problem, can anyone answer it and show working , so I can learn. Thanks

Given P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6 find:

(a) P(R) (b) P(S|R) (c) P(S'|R) (d) P(S'|R')

2. P(R)
= P(R(S+S'))
= P(R1)
= P(RS)+P(RS')
= P(R|S)P(S)+P(R|S')P(S')
= P(R|S)P(S)+P(R|S')(1-P(S))

P(RS)
= P(R|S)P(S)
= P(S|R)P(R)

Now you know P(R) and P(RS), so you know P(RS')
P(RS')
= P(S'|R)P(R)

P(S)
= ...
= P(S|R)P(R)+P(S|R')(1-P(R))
gives
P(SR')
= P(S|R')P(R')
= P(S|R')(1-P(R))
Using this,
P(1)
= P((R+R')(S+S'))
= P(RS) + P(RS')+P(R'S)+P(R'S')
gives
P(R'S')
= P(S'|R')P(R')

3. Originally Posted by kamicool17
Given P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6 find:
(a) P(R) (b) P(S|R) (c) P(S'|R) (d) P(S'|R')
$P(RS)=P(R|S)P(S)$
$P(RS')=P(R|S')P(S')$
$P(R)=P(RS)+P(RS')$

Now finish.

4. Hello, kamicool17!

Another approach . . . (much longer, though)

Given: . $P(R|S) = 0.5,\;\;P(R|S') = 0.4,\;\;P(S) = 0.6$

Find: . $(a)\;P(R) \qquad (b)\;P(S|R)\qquad (c)\;P(S'|R) \qquad (d)\;P(S'|R')$

We have this table:

. . $\begin{array}{c|| c|c||c}
& S & S' & \text{Total } \\ \hline \hline
R & {\color{blue}(a)} & {\color{blue}(c)} & {\color{blue}(e)} \\ \hline
R' & {\color{blue}(b)} & {\color{blue}(d)} & {\color{blue}(f)} \\ \hline \hline
\text{Total} & 0.60 & 0.40 & 1.00 \end{array}$

We have: . $P(S) = 0.60 \quad\Rightarrow\quad P(S') = 0.40$
So we can fill in the bottom row.

We have: . $P(R|S) = 0.5$

Then: . $\frac{P(R \wedge S)}{P(S)} \;=\;0.5 \quad\Rightarrow\quad \frac{P(R \wedge S)}{0.6} \:=\:0.5
$

Hence: . $P(R \wedge S) \,=\,0.30\;{\color{blue}(a)} \quad\Rightarrow\quad P(R' \wedge S) \:=\:0.30\;{\color{blue}(b)}$

. . $\begin{array}{c|| c|c||c}
& S & S' & \text{Total } \\ \hline \hline
R & 0.30 & (c) & (e) \\ \hline
R' & 0.30 & (d) & (f) \\ \hline \hline
\text{Total} & 0.60 & 0.40 & 1.00 \end{array}$

We have: . $P(R|S') = 0.4$

Then: . $\frac{P(R \wedge S')}{P(S')} \:=\:0.4 \quad\Rightarrow\quad \frac{P(R \wedge S')}{0.4} \:=\:0.4$

Hence: . $P(R \wedge S') \:=\:0.16\;{\color{blue}(c)} \quad\Rightarrow\quad P(R' \wedge S') \:=\:0.24\;{\color{blue}(d)}$

Also: . $P(R) \:=\:0.46 \;{\color{blue}(e)}\quad\Rightarrow\quad P(R') \:=\:0.54\;{\color{blue}(f)}$

And we have completed the table:

. . $\begin{array}{c|| c|c||c}
& S & S' & \text{Total } \\ \hline \hline
R & 0.30 & 0.16 & 0.46 \\ \hline
R' & 0.30 & 0.24 & 0.54 \\ \hline \hline
\text{Total} & 0.60 & 0.40 & 1.00 \end{array}$

You can now answer the questions . . .