Answer:
P(C4) = 0.0711
Step-by-step explanation:
consider the first draw = 15/23 since it cannot be a blue ball
The second draw = 21/29 since 6 more red balls will be added after the draw since a blue ball cannot be drawn
the third draw = 27/35 since 6 more red balls will be added after each draw since a blue ball cannot be drawn
therefore the total number of red balls will be = 15 + 6 + 6 + 6 = 33 red balls after the 4th draw. the total ball now in the urn= 33 red + 4 blue = 41
Hence the probability of drawing a blue ball at the fourth draw after drawing red balls at the previous attempts = 8/41
P(C4) = P ( fourth ball is blue ) * P( first ball red)*P(second ball red) *P(third ball red )
= (8/41) * (15/23) * (21/29)* (27/35) = 0.0711
11=y/6
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hope it helps
sorry, I didn't know how to explain it
I believe that this would be representative.
