Question

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Two balls are drawn in succession out of a box containing 2 red and 4 white balls. Find the probability that at least 1 ball was red, given that the first ball was (A) Replaced before the second draw. (B) Not replaced before the second draw. (A) Find the probability that at least 1 ball was red, given that the first ball was replaced before the second draw. (Simplify your answer. Type an integer or a fraction.) (B) Find the probability that at least 1 ball was red, given that the first ball was not replaced before the second draw. (Simplify your answer. Type an integer or a fraction)

Answer 1

Q. Solution : CA) We define two events As deeswuſ a wlite ball mithe Bst selection Be drawing a white ball on the end eselection PCA)= 놓 =를 " surce there are 4 while balls and total mot balls = 4+2=6 there is replacement, So, PCB/A)= P(B)= P(A), Wlucha có borrible cbecause of the enidefensent events foccused on the same colour for and because we au each event. PC A and B)= PCA) + P(BIA) = P(A) * P(B) 16 We have to fund prebab Elity at least I ball 8/+ 2/3 Lolf was need then plat least one red ) = d - P(no reeds) -t-P(A and B 11 1- 4/9 = 9-4 5/ 0.5556 =55.56% (B) Now, PLB/A) = P(B) is not true due to the fact that replacement is that made. Suice P(B) = 46 = 45 ad replacement aimade but now P(B) = 4-1 6-1 8(8)= 375

therefore P(BIA)= 3/5 مر esor PLA and B)= PLA)* ple/x) = 4%* *s are res als cind plat cost one oud) - s-Pl noreda) = I-P,itwo Colude) 0:1-4 = 2 60l be cdoletion. 3/8 20.6
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