Respuesta :
Answer:
(A) 0.04
(B) 0.25
(C) 0.40
Step-by-step explanation:
Let R = drawing a red chips, G = drawing green chips and W = drawing white chips.
Given:
R = 8, G = 10 and W = 2.
Total number of chips = 8 + 10 + 2 = 20
[tex]P(R) = \frac{8}{20}=\frac{2}{5}\\P(G)= \frac{10}{20}=\frac{1}{2}\\P(W)= \frac{2}{20}=\frac{1}{10}[/tex]
As the chips are replaced after drawing the probability of selecting the second chip is independent of the probability of selecting the first chip.
(A)
Compute the probability of selecting a white chip on the first and a red on the second as follows:
[tex]P(1^{st}\ white\ chip, 2^{nd}\ red\ chip)=P(W)\times P(R)\\=\frac{1}{10}\times \frac{2}{5}\\ =\frac{1}{25} \\=0.04[/tex]
Thus, the probability of selecting a white chip on the first and a red on the second is 0.04.
(B)
Compute the probability of selecting 2 green chips:
[tex]P(2\ Green\ chips)=P(G)\times P(G)\\=\frac{1}{2} \times\frac{1}{2}\\ =\frac{1}{4}\\ =0.25[/tex]
Thus, the probability of selecting 2 green chips is 0.25.
(C)
Compute the conditional probability of selecting a red chip given the first chip drawn was white as follows:
[tex]P(2^{nd}\ red\ chip|1^{st}\ white\ chip)=\frac{P(2^{nd}\ red\ chip\ \cap 1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\=\frac{P(2^{nd}\ red\ chip)P(1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\= P(R)\\=\frac{2}{5}\\=0.40[/tex]
Thus, the probability of selecting a red chip given the first chip drawn was white is 0.40.
