1. ## Word problem

I need some help and clarifications on such problems. For example:

A student thinks he will succeed on subjects "Probability", "Statistics" and in both of them with the probabilities 0.7, 0.6 and 0.5 respectively. Find the probabilities:
1). p1: The student will succeed at least in one of the subjects
2), p2: The student will succeed only in one of the subjects
3), p3: The student will succeed not more than in one subject
result: p1 = p2+p3

2. Originally Posted by javax
I need some help and clarifications on such problems. For example:

A student thinks he will succeed on subjects "Probability", "Statistics" and in both of them with the probabilities 0.7, 0.6 and 0.5 respectively. Find the probabilities:
1). p1: The student will succeed at least in one of the subjects
2), p2: The student will succeed only in one of the subjects
3), p3: The student will succeed not more than in one subject
$p_1=P(P)+P(S)-P(B)$
$p_2=P(P)+P(S)-2P(B)$
$p_3=1-P(B)$

3. Originally Posted by Plato
$p_1=P(P)+P(S)-P(B)$
$p_2=P(P)+P(S)-2P(B)$
$p_3=1-P(B)$
ok thanks, would you please explain how you derived formulas for p2 and p3?

4. Originally Posted by javax
ok thanks, would you please explain how you derived formulas for p2 and p3?
Exactly one:
$p_2=P\left( {A \cap B^c } \right) + P\left( {A^c \cap B} \right) =$ $\left[ {P(A) - P\left( {A \cap B} \right)} \right] + \left[ {P(B) - P\left( {A \cap B} \right)} \right] = \left[ {P(A) + P(B) - 2P\left( {A \cap B} \right)} \right]$

$p_3$ is "No more than one" is "not both".