Question

A bag contains 6 red apples and 8 green apples. You choose two apples. What is the probability of picking a red and a green apple without replacement?

Answer 1

Take consideration about whether this event is dependent or independent. If we pick an apple of a particular colour in the first draw it will not affect the outcomes for those apples of another colour. So the outcome of the first choice will never affect the outcome of the second choice, thus this draw of apples without replacement becomes an independent event.

Complete step-by-step answer:
Probability is defined as the number of favourable outcomes (F) divided by the total number of outcomes (T). We have a total of 14 apples out of which green ones are 6 and red ones are 8.
$\bullet$ $P\left( E \right) = \dfrac{F}{T}$
Let’s pick a green apple in the first draw:
$\bullet$ Probability of getting a green apple in the first draw, $P\left( G \right) = \dfrac{6}{{14}} = \dfrac{3}{7}$
Now after getting the apple we did not replace it,
$\bullet$ So total number of apples remain $= 13$
$\bullet$ Probability of getting a red apple in the second draw, $P\left( R \right) = \dfrac{8}{{13}} = \dfrac{8}{{13}}$
Since both of them are independent events, we know that for independent events:
$\bullet$ $P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)$
Now our draw of apples is also independent so probability of getting a green and a red apple without replacement:
$\bullet$ $P\left( {G \cap R} \right) = P\left( G \right) \times P\left( R \right)$
$\bullet$ Substituting the values for $P\left( R \right)$ and $P\left( G \right)$ in the above equation
$\bullet$ $P\left( {G \cap R} \right) = \dfrac{3}{7} \times \dfrac{8}{{13}}$
$\bullet$ $P\left( {G \cap R} \right) = \dfrac{{24}}{{91}}$
Now if you want to reverse the order by picking a red apple first then the outcome will remain the same as the events are independent to each other.
Let’s pick a red apple in the first draw:
$\bullet$ Probability of getting a green apple in the first draw, $P\left( G \right) = \dfrac{8}{{14}} = \dfrac{4}{7}$
Now after getting the apple we did not replace it,
$\bullet$ So total number of apples remain $= 13$
$\bullet$ Probability of getting a green apple in the second draw, $P\left( R \right) = \dfrac{6}{{13}} = \dfrac{6}{{13}}$
Now our draw of apples is also independent so probability of getting a green and a red apple without replacement:
$\bullet$ $P\left( {G \cap R} \right) = P\left( G \right) \times P\left( R \right)$
$\bullet$ Substituting the values for $P\left( R \right)$ and $P\left( G \right)$ in the above equation
$\bullet$ $P\left( {G \cap R} \right) = \dfrac{4}{7} \times \dfrac{6}{{13}}$
$\bullet$ $P\left( {G \cap R} \right) = \dfrac{{24}}{{91}}$
We tried from both the ways and the outcomes remain the same which is $\dfrac{{24}}{{91}}$as both the events are independent.

Note: Before jumping into these kinds of problems which involve probability or permutation and combination. It is very necessary to understand the problem and try to figure out exactly the type of event taking place, whether the type of event is a dependent event or independent event.once the proper type of event is analyzed the question and the solving process becomes very clear to us in this way.