# A simple propability question

• Jan 8th 2007, 10:14 AM
jimmybob
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!
• Jan 8th 2007, 10:28 AM
ThePerfectHacker
Quote:

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
Quote:

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
• Jan 8th 2007, 11:22 AM
Soroban
[soze=3]Hello, jimmybob![/size]

Quote:

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

(1) Find the probability that both will fail

(2) Find the probability that only one will pass

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

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

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

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

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

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

• Jan 8th 2007, 11:27 AM
Plato
Quote:

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: $\displaystyle \left( {1 - \frac{3}{5}} \right)\left( {\frac{4}{7}} \right).$

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