Answer:
Dependent
Step-by-step explanation:
Given that there are 15 balls in an urn. 7 of the balls are red, and 8 of the balls are blue
when we do with replacement each time probability for red or blue remains the same.
a) the probability that if you draw one ball, it will be blue=[tex]\frac{8}{15}[/tex]
(b) If you draw two balls without replacement, the probability that you draw a red and then a blue
=[tex]\frac{7}{15} *\frac{8}{14} \\=\frac{4}{15}[/tex]
(c) If you draw two balls with replacement, the probability that you draw a red and then a blue
=[tex]\frac{7}{15} *\frac{8}{15} \\=\frac{56}{225}[/tex]
(d) If you draw three balls without replacement, the probability that you draw at least two red balls
=P(2 red, 1 blue)+P(3 blue)
=[tex]\frac{7C2*8C1}{15C3} +\frac{7C3}{15C3} \\=\frac{203}{455} =\frac{29}{65}[/tex]
(e) If you draw three balls with replacement, the probability that you draw at least two red balls
= P(X≥2) where x is binomial with n=3 and p = 7/15
= 0.4506
(f) We play a game where I draw a ball first without replacement, and then you draw one. I draw a blue ball, and then you draw a red ball.
No when first ball is drawn without replacement the next probability for blue and red would be changed. So dependent.
