This problem is familiarly known as the "probability of independent events."
a. Let [tex]A=[/tex] the event that the first ball is blue, and [tex]B=[/tex] the event that the second ball is blue.
In the beginning, there are 35 balls in the box, 12 of which are blue. Therefore, expressing it into a probability form as [tex]P(A)= \frac{12}{35} [/tex].
After the first selection, the ball was returned to the box making the number of balls in the box still equal to 12 blue, 14 red, and 9 yellow balls. So [tex]P(B|A)= \frac{12}{35} [/tex].
Thus, multiplying the two probabilities yields
[tex]P_A_a_n_d_B= \frac{12}{35}( \frac{12}{35} )= \frac{144}{1225}=0.11755 [/tex]
b. Let [tex]A=[/tex] the event that the first ball is red, and [tex]B=[/tex] the event that the second ball is yellow.
Again there are 35 balls, 14 of which are red: [tex]P(A)= \frac{14}{35} [/tex]
For the second selection to be a yellow ball (as which the returning of the first ball to the box is allowed), [tex]P(B|A)= \frac{9}{35} [/tex]
Thus, [tex]P_A_a_n_d_B= \frac{14}{35}( \frac{9}{35} )= \frac{18}{175}=0.10286 [/tex]
