A bag contains 6 red marbles, 6 white marbles, and 4 blue marbles. What is the probability that someone could pull a white marble from the bag, keep it, and then pull a red marble on the second try?

Answer:The probability of getting white marble in the first try and red marble in the second try (without replacement) is [tex]\frac{3}{20}[/tex]Step-by-step explanation:Given: A bag contains 6 red marbles, 6 white marbles, and 4 blue marbles.The total numb[tex][/tex]er of marbles in the bag = 6 + 6 + 4 = 16Someone is pull a marble.The probability of getting a white marble P(W)= [tex]\frac{The number of favorable outcomes}{The total number of possible outcomes}[/tex]P(W) = [tex]\frac{6}{16}[/tex]Then that marble kept it (with out replacement).Now the total number of marbles in the bag = 6(red) + 5(white) + 4(blue) = 15In the second try, the probability of getting a red marble P(R) = [tex]\frac{6}{15}[/tex]These two events are dependent, because the out comes of the first event influences the outcome of the second event.So, the probability that someone could pull a white marble from the bag, keep it, and then pull a red marble on the second try = P(W) *P(R)= [tex]\frac{6}{16} *\frac{6}{15}[/tex]Now we can multiply the fractions and simplify.= [tex]\frac{36}{240}[/tex]Here GCF of 36 and 240 is 12. So divide both the numerator and the denominator by 12, we get= [tex]\frac{3}{20}[/tex]So, the probability of getting white marble in the first try and red marble in the second try (without replacement) is [tex]\frac{3}{20}[/tex]