# Probability help (this is driving me nuts) #3

• September 16th 2009, 11:11 AM
SydneyBristow
Probability help (this is driving me nuts) #3
I have been at this for 3 days already and i did some of them but i just am so confused that my brain hurts even trying to figure out if i am doing the right stuff. please please help me/ show me or point me in the right direction with these problems

1.English and American spellings are colour and color, respectively. A man staying at a
Parisian hotel writes this word, and a letter taken at random from his spelling is found to be a vowel. If
40 percent of the English-speaking men at the hotel are English and 60 percent are Americans, what is the
probability that the writer is an Englishman?

4. Three missiles are fired at a target and hit it independently with probabilities 0.6, 0.7, and
0.8, respectively. What is the probability that the target is hit?
• September 16th 2009, 03:14 PM
Soroban
Hello, SydneyBristow!

Here's the first one . . .

Quote:

1. English and American spellings are "colour" and "color", respectively.
A man staying at a Parisian hotel writes this word, and a letter taken
at random from his spelling is found to be a vowel.
If 40% of the English-speaking men at the hotel are English and 60% are Americans,
what is the probability that the writer is an Englishman?

This requires repeated use of Bayes' Theorem . . .

We have: . $\begin{array}{cccc}P(\text{UK}) \:=\: 0.4 & & P(\text{vowel}\,|\,\text{UK}) \:=\: \frac{3}{6} \:=\:0.5 & {\color{blue}[1]} \\ \\[-4mm] P(\text{US}) \:=\:0.6 & & P(\text{vowel}\,|\,\text{US}) \:=\:\frac{2}{5} \:=\:0.4& {\color{blue}[2]}\end{array}$

We want: . $P(\text{UK}\,|\,\text{vowel}) \;=\;\frac{P(\text{UK} \wedge\text{vowel})}{P(\text{vowel})} \;\;{\color{blue}[3]}$

From [1] we have: . $\frac{P(\text{vowel}\wedge\text{UK})}{P(\text{UK}) } \:=\:0.5 \quad\Rightarrow\quad \frac{P(\text{vowel}\wedge\text{UK})}{0.4} \:=\:0.5$

. . $P(\text{vowel}\wedge\text{UK}) \:=\:(0.4)(0.5) \:=\:0.20\;\;{\color{blue}[4]}$

From [2] we have: . $\frac{P(\text{vowel}\wedge\text{US})}{P(\text{US}) } \;=\; 0.4 \quad\Rightarrow\quad \frac{P(\text{vowel}\wedge\text{US})}{0.6} \;=\;0.4$

. . $P(\text{vowel}\wedge\text{US}) \;=\;(0.6)(0.4) \;=\;0.24$

Hence: . $P(\text{vowel}) \;=\;P(\text{vowel}\wedge\text{UK}) + P(\text{vowel}\wedge\text{US}) \;=\;0.20 + 0.24 \;=\;0.44\;\;{\color{blue}[5]}$

Substitute [4] and [5] into [3]: .

. . $P(\text{UK}\,|\,\text{vowel}) \;=\;\frac{P(\text{UK}\wedge\text{vowel})}{P(\text {vowel})} \;=\;\frac{0.20}{0.44} \;=\;\frac{5}{11}$

• September 16th 2009, 04:29 PM
SydneyBristow
thanks so much
thankyou .

can you plz advise if for the missle problem my logic accurate?

1-(.4*.3*.2 )=.976

i cant type much as i hv newborn in my arns
• September 16th 2009, 04:40 PM
artvandalay11
Probability that at least 1 hits = 1-Probability that every one misses

=1-(.4*.3*.2) as you said, looks good to me