1. ## statistics

can you help me with the quetion number 4 please and may be 6 too!

http://www.robinson.homechoice.co.uk...on%20Paper.pdf

thanks

can you help me with the quetion number 4 please and may be 6 too!

http://www.robinson.homechoice.co.uk...on%20Paper.pdf

thanks
Q4 looks like a routine application of the calculational formulae. Do you have examples to follow?

Q6: Pr(H) = 5/8, Pr(W) = 1/2, Pr(H | W) = 5/6.

(a) Pr(H | W) = Pr(H and W)/Pr(W). Substitute the given data and re-arrange to solve for Pr(H and W).

(c) 1 - Pr(H and W) (why?).

(d) Pr(H' and W') = 1 - Pr(H or W). Or just get it straight off Venn diagram.
Note: Pr(H or W) = Pr(H) + Pr(W) - Pr(H and W).

(e) The required events are $H_1 ~ W_1 ~ H_2' ~ W_2' ~$ or $~ H_1' ~ W_1 ~ H_2 ~ W_2' ~$ or ....... (there are more - do you see them?). Now calculate the probability of each. Add these probabilities together.

Here's #6 . . .

6. For any married couple, the probability that the husband passed his driving test is $\frac{5}{8}$
and the prob. that the wife passed her driving test is $\frac{1}{2}$
The prob. that the husband has passed, given that his wife has passed, is $\frac{5}{6}$
A married couple is chosen random.
(a) Show that the probability that both of them have passed is $\frac{5}{12}$
We are given: . $P(H) = \frac{5}{8},\;P(W) = \frac{1}{2},\;P(H|W) = \frac{5}{6}$

Bayes' Theorem: . $P(H|W) \;=\;\frac{P(H \wedge W)}{P(W)}$

Substitute: . $\frac{5}{6} \;=\;\frac{P(H \wedge W)}{\frac{1}{2}} \quad\Rightarrow\quad P(H \wedge W) \;=\;\frac{5}{6}\!\cdot\!\frac{1}{2}$

Therefore: . $P(H \wedge W) \;=\;\frac{5}{12}$

b) Draw a Venn diagram to represent these data.
Code:
  * - - - - - - - - - - - - - - - - - - - *
|                                       |
|       * - - - - - - - *               |
|       | H             |               |
|       |       * - - - + - - - *       |
|       |  5/24 |       |       |       |
|       |       |  5/12 |       |       |
|       |       |       | 1/12  |       |
|       * - - - + - - - *       |       |
|               |             W |       |
|               * - - - - - - - *       |
|                                 7/24  |
* - - - - - - - - - - - - - - - - - - - *
Find the probability that:

c) only one of them has passed their driving test.
$P(H \veebar W) \;=\;\frac{5}{24} + \frac{1}{12} \;=\;\frac{7}{24}$

d) neither of them has passed their driving test.
$P(H' \wedge W') \;=\;\frac{7}{24}$

Two couples are chosen at random.
e) Find the proability that only one of the two husbands and
only one of the two wives have passed their driving test.
One husband passed and the other failed.
This can happen in two ways:
. . $P(H \wedge H') \:=\:\frac{5}{8}\!\cdot\!\frac{3}{8}\;=\;\frac{15} {64} \;\;\text{ or }\;\;P(H' \wedge H) \:=\:\frac{3}{8}\!\cdot\!\frac{5}{8} \:=\:\frac{15}{64}$
Hence: . $P(\text{one H passed}) \:=\:\frac{15}{64}+\frac{15}{64} \:=\:\frac{15}{32}$

One wife passed and the other failed.
This can happen in two ways:
. . $P(W \wedge W') \:=\:\frac{1}{2}\!\cdot\!\frac{1}{4}\;\;\text{ or }\;\;P(W' \wedge W) \:=\:\frac{1}{2}\!\cdot\!\frac{1}{2}\:=\:\frac{1}{ 4}$
Hence: . $P(\text{one W passed}) \:=\:\frac{1}{4} + \frac{1}{4} \:=\:\frac{1}{2}$

Therefore: . $P(\text{one H passed }\wedge\text{ one W passed}) \;=\;\frac{15}{32}\!\cdot\frac{1}{2} \;=\;\frac{15}{64}$

4. ## statistics

can any one help on quetion number 5 plz

http://www.robinson.homechoice.co.uk...on%20Paper.pdf

Thanks