1. A simple propability question

Two learners are due to take an exam, The probability that Andy will pass is 3/5 and the probability that Brenda will fail is 4/7

first i need to calculate the probability that both will fail

then i need the probability that only one will pass

and..

If brenda fails, the probability that she will pass next time is 2/5
a) calculate the probability that brenda fails again
b) calculate the probability brenda passes having taken the exam for the second time.

Any help is much appreciated!

2. Originally Posted by jimmybob
Two learners are due to take an exam, The probability that Andy will pass is 3/5 and the probability that Brenda will fail is 4/7

first i need to calculate the probability that both will fail
First will fail is 1-3/5=2/5
Second will fail is 1-4/7=3/7
Thus, we multiply the probabilities,
(2/5)(3/7)=6/35
then i need the probability that only one will pass
2 possibilities...
1)First person fails and second person passes.
2)First person passes and first person fails.
The probability in #1 is:
(2/5)(4/7)=8/35
The probability in #2 is:
(3/5)(3/7)=9/35
This is an OR question. Meaning #1 OR #2 thus we add,
8/35+9/35=17/35

3. [soze=3]Hello, jimmybob![/size]

Andy and Brenda take an exam,
The probability that Andy passes is $\frac{3}{5}$ and the probability that Brenda fails is $\frac{4}{7}$.

(1) Find the probability that both will fail

(2) Find the probability that only one will pass

We have these individual probabilities:
. . $\begin{array}{cc}P(\text{Andy passes}) = \frac{3}{5} & P(\text{Andy fails}) = \frac{2}{5} \\ P(\text{Brenda passes}) = \frac{3}{7} & P(\text{Brenda fails}) = \frac{4}{7} \end{array}$

(1) $P(\text{both fail}) \:=\:P(\text{Andy fails and Brenda fails}) \:=\:\frac{2}{5}\cdot\frac{4}{7}\:=\:\boxed{\frac{ 8}{35}}$

(2) If exactly one of them passes, there are two cases:

. . Andy passes and Brenda fails: . $P(A^+\text{ and }B^-) \:=\:\frac{3}{5}\cdot\frac{4}{7} \:=\:\frac{12}{35}$

. . Andy fails and Brenda passes: . $P(A^-\text{ and }B^+) \:=\:\frac{2}{5}\cdot\frac{3}{7}\:=\:\frac{6}{35}$

. . Hence: . $P(\text{one passes}) \:=\:\frac{12}{35} + \frac{6}{35}\:=\:\boxed{\frac{18}{35}}$

4. Originally Posted by jimmybob
The probability that Andy will pass is 3/5 and
the probability that Brenda will fail is 4/7.
The probability that both will fail: $\left( {1 - \frac{3}{5}} \right)\left( {\frac{4}{7}} \right).$

The probability that only one will pass: $\left( {\frac{3}{5}} \right)\left( {\frac{4}{7}} \right) + \left( {\frac{2}{5}} \right)\left( {\frac{3}{7}} \right).$