A box contains 15 milk chocolates and 5 plain chocolates. Two chocolates are chosen at random. Calculate the probability that

#= P["1st is milk AND 2nd is plain"]+P["1st is plain AND 2nd is milk"]#
#= (15/20)(5/19) + (5/20)(15/19)#
#= 2*(15/20)(5/19)#
#= 2*(3/4)(5/19)#
#= (3/2)(5/19)#
#= 15/38#
#= 0.3947#
#= 39.47 %#

#"Explanation : "#
#"When we first pick one there are 20 chocolates in the box."#
#"When we pick one after that, there are 19 chocolates in the box."#
#"We use the formula"#
#P[A and B] = P[A] * P[B|A]#
#"because both draws are not independent."#
#"So take e.g. A = '1st is milk' and B = '2nd is chocolate'"#
#"Then we have"#
#P[A]= 15/20 " (15 milks on 20 chocolates)"#
#P[B|A] = 5/19#
#"(5 plain left on 19 chocs in total left after drawing a milk at first)"#

Quick way to visualize this kind of probability question:

Suppose we have a bag of #N# marbles of many different colours, and we're interested in the probability of selecting

#n_1# out of #N_1# red marbles
#n_2# out of #N_2# yellow marbles
...
#n_k# out of #N_k# purple marbles

where the sum of all the #n_i"'s"# is #n# and the sum of all the #N_i"'s"# is #N.#

Then the probability is equal to:

#[((N_1),(n_1))((N_2),(n_2))...((N_k),(n_k))]/(((N),(n)))#

For this question, the formula becomes:

#[((15),(1))((5),(1))]/[((20),(2))]#

which is equal to

#[" "15 xx 5" "]/[(20xx19)/(2xx1)]=75/190=15/38~~39.5%#