according to the general equation for conditional probability of p(an B) = 3/10 and P(B)= 3/5, what is P(A|B)

2 answers:

4 0

The required formula is:
$P(A | B)=\frac{P(A\cap B)}{P(B)}$
Therefore the probability of A given B is:
$P(A|B)=\frac{\frac{3}{10}}{\frac{3}{5}}=\frac{3\times5}{10\times3}=\frac{1}{2}$

astraxan [27]3 years ago

4 0

Answer: its 1/2

Step-by-step explanation:

You might be interested in

I need help with this

andrey2020 [161]

18a:16 18b:32 % means out of 100

3 0

3 years ago

I need help I can’t figure it out

dusya [7]

Answer:

$\large\boxed{\dfrac{41p}{70q^4(r-9)^2}}$

Step-by-step explanation:

$\dfrac{2p^4}{5q^5(r-9)^3}\cdot\dfrac{41q(r-9)}{28p^3}\\\\=\dfrac{2\!\!\!\!\diagup^1\cdot41}{5\cdot28\!\!\!\!\!\diagup_{14}}\cdot\dfrac{q\!\!\!\!\diagup}{q^{5\!\!\!\!\diagup^4}}\cdot\dfrac{(r\!\!\!\!\!\!{--}-9)\!\!\!\!\!\!{--}}{(r-9)^{3\!\!\!\!\diagup^2}}\cdot\dfrac{p^{4\!\!\!\!\diagup}}{p^3\!\!\!\!\!\!\diagup}\\\\=\dfrac{41}{70}\cdot\dfrac{1}{q^4}\cdot\dfrac{1}{(r-9)^2}\cdot\dfrac{p}{1}\\\\=\dfrac{41p}{70q^4(r-9)^2}$