Question

1. An urn initially contains 6 red and 8 green balls. Each time a ball is selected, its color is recorded, and it is replaced in the urn along with 2 other balls of the same color. Compute the probability that:

(a) The first 2 balls selected are green and the next 2 are red?

(b) Of the first 4 balls selected, exactly 2 are green?

(c) If the second ball selected is green, what is the probability that the first one was red?

1. An urn initially contains 6 red and 8 green balls. Each time a ball is selected, its color is recorded, and it is replaced in the urn along with 2 other balls of the same color. Compute the probability that: (a) The first 2 balls selected are green and the next 2 are red? (b) Of the first 4 balls selected, exactly 2 are green? (c) If the second ball selected is green, what is the probability that the first one was red?

Answer 1

(a) the first 2 balls selected are green and the next 2 are red

Let two letters G and R for GREEN and RED respectfully.

we have to find P(GGRR) = P(GGRR/GGR).P(GGR/GG).P(GG/G).P(G) ( . Shows MULTIPLICATION)

P(G) = 8/14 (8 GREEN AND 6 RED)

P(GG/G) = 10/16 (10 GREEN AND 6 RED)

P(GGR/GG) = 6/18 (6 RED AND 12 GREEN)

P(GGRR/GGR)= 8/20 (8 RED AND 12 GREEN)

So P(GGRR) = 8/20 * 6/18 * 10/16 * 8/14

finally, P(GGRR) =3/63 ANSWER

(b) Of the first 4 balls selected, exactly 2 are green

Using Same Process of Part we find this

P(GGRR)+P(GRGR)+P(GRRG)+P(RGRG)+P(RGGR)+P(RRGG)

=4*3/63 = 12/63 = 0.1905 Approx ANSWER (Each Function have Value 3/63)

(c) If the second ball selected is green, what is the probability that the first one was red

Similarly From Part a and b it can be Solved

we have to find P(RG) = 6/14 * 8/16

=3/14 ANSWER