Answer:
Generally the probability distribution of the number of white balls (X) is
X 0 1 2
P(X) 0.35714 0.5357 0.10714
Step-by-step explanation:
From the question we are told that
The number of white balls is [tex]N_w = 5[/tex]
The number of black balls is [tex]N_b = 3[/tex]
Generally the total number of balls the urn contains is
[tex]N_t = N_w + N_b = 5 + 3 = 8 \ balls[/tex]
Generally the probability that of the two balls drawn that the balls are 2 white balls and 0 black balls is
[tex]P( 2_w \ , 0_b) = \frac{ ^{N_w} C _2 * ^{N_b} C _0 }{^{N_t} C_2}[/tex]
Here C stands for combination hence we would be making use of the combination feature in our calculator
=> [tex]P( 2_w \ , 0_b) = \frac{ ^5 C _2 * ^3 C _0 }{^{8} C_2}[/tex]
=> [tex]P( 2_w \ , 0_b) = 0.10714[/tex]
Generally the probability that of the two balls drawn that the balls are 1 white balls and 1 black balls is
[tex]P( 1_w \ , 1_b) = \frac{ ^{N_w} C _1 * ^{N_b} C _1 }{^{N_t} C_2}[/tex]
=> [tex]P( 1_w \ , 1_b) = \frac{ ^5 C _1 * ^3 C _1 }{^{8} C_2}[/tex]
=> [tex]P( 1_w \ , 1_b) = 0.5357[/tex]
Generally the probability that of the two balls drawn that the balls are 0 white balls and 2 black balls is
[tex]P( 0_w \ , 2_b) = \frac{ ^{N_w} C _0 * ^{N_b} C _2}{^{N_t} C_2}[/tex]
=> [tex]P( 0_w \ , 2_b) = \frac{ ^5 C _0 * ^3 C _2 }{^{8} C_2}[/tex]
=> [tex]P( 0_w \ , 2_b) = 0.35714[/tex]
Generally the probability distribution of the number of white balls (X) is
X 0 1 2
P(X) 0.35714 0.5357 0.10714
