# Thread: [SOLVED] Set probability questions

1. ## [SOLVED] Set probability questions

1 ] The events A and B are such that
P(A) = 0.2, P(B) = 0.6 and P(A|B) = 0.3.
Find
(a) P(A∩B), [2]
(b) P(B|A), [3]
(c) P(A∪B). [3]
(d) P(A‘∩B‘). [2]

2] The events A and B are such that P(A) = 0.5 and P(A∪B) = 0.7. Determine the value of P(B) in each of the cases when
(a) A and B are mutually exclusive, [2]
(b) A and B are independent, [4]
(c) P(B|Α) = 0.3. [3]

1]
a] 0.18
b] 0.9
c] 0.62
d] 0.82

2]
a] 0.2
b] Cannot understand how to work out

Sorry I couldn't post working, I have it all written down and it was difficult enough to write all of the questions down! If I am wrong, could you please give me a clue towards getting it right and could you please explain how to attempt 2b]?

2. Hello, db5vry!

As we say in The Corps, ya done good!

The events $A$ and $B$ are such that

$P(A) = 0.2,\;\;P(B) = 0.6,\;\;P(A|B) = 0.3$

Find:

$(a)\;P(A\cap B)\quad{\color{blue}0.18}\quad{\color{red} \hdots \text{ Yes!}}$

$(b)\;P(B|A)\quad{\color{blue}0.9}\quad{\color{red} \hdots \text{ Right!}}$

$(c)\;P(A \cup B)\quad{\color{blue}0.62}\quad {\color{red} \hdots \text{ Correct!}}$

$(d)\:P(A' \cap B')\quad{\color{blue}0.82}\quad{\color{red}\hdots \text{ no}}$

$A' \cap B'$ is the complement of $A\cup B$

So: . $P(A' \cap B') \:=\:1 - 0.62 \:=\:\boxed{0.38}$

2] The events $A$ and $B$ are such that: $P(A) = 0.5,\;P(A \cup B) = 0.7$

Determine the value of $P(B)$ in each of the cases when:

(a) $A$ and $B$ are mutually exclusive. .0.2 . . . . Right!

(b) $A$ and $B$ are independent.

(c) $P(B|A) = 0.3$

(b) If $A$ and $B$ are independent: . $P(A \cap B) \:=\:P(A)\!\cdot\!P(B) \:=\:(0.5)(0.7) \:=\:0.35$

Substitute into: . $P(A \cup B) \:=\:P(A) + P(B) - P(A \cap B)$

. . . . . . . . . . . . . . . $0.7 \quad=\;0.5 \;+ \;P(B)\quad -\quad 0.35$

. . . . . . . . . . . . . . $P(B) \;=\;0.7 - 0.5 + 0.35 \;=\;\boxed{0.55}$

(c) $P(B|A) = 0.3$

From Bayes' Theorem: . $P(B|A) \:=\:\frac{P(B \cap A)}{P(A)} \:=\:0.3$

So we have: . $\frac{P(A \cap B)}{0.5} \:=\:0.3 \quad\Rightarrow\quad P(A \cap B) \:=\:(0.3)(0.5) \:=\:0.15$

Substitute into: . $P(A \cup B) \:=\:P(A) + P(B) - P(A \cap B)$

. . . . . . . . . . . . . . . $0.7 \quad=\;0.5 \;+ \;P(B)\quad -\quad 0.15$

. . . . . . . . . . . . . . $P(B) \;=\;0.7 - 0.5 + 0.15 \;=\;\boxed{0.35}$