Posted 15 Apr 2014 · Report post Five pieces are randomly selected from a standard chess set. If it is known that at least one of these pieces is a pawn, what is the probability that at least two of these five pieces are white pawns? 0 Share this post Link to post Share on other sites

0 Posted 16 Apr 2014 · Report post is the answer 3874 / (32c5) ?? 0 Share this post Link to post Share on other sites

0 Posted 16 Apr 2014 · Report post Let A be the event that at least two white pawns are selected, and let B be the event that at least one pawn is selected. We seek P(A|B). Since A implies B, we have P(A and B) = P(A). So P(A|B) = P(A and B)/P(B) = P(A)/P(B). So we just need to calculate P(A) and P(B), which is fairly straightforward. P(A) = 1-P(less than two white pawns selected) = 1-P(exactly one white pawn selected)-P(no white pawns selected) = 1-24*23*22*21*40/32^{5}-24*23*22*21*20/32^{5} = 1-24*23*22*21*60/32^{5} = 1-23*11*7*5*3^{3}/2^{19} = 285203/524288 P(B) = 1-P(no pawns selected) = 1-16*15*14*13*12/32^{5} = 1-13*7*5*3^{2}/2^{18} = 258049/262144 P(A)/P(B) = 285203/516098, which would be the probability we seek. 0 Share this post Link to post Share on other sites

0 Posted 21 Apr 2014 · Report post there are 8 pawns, 2 rooks, 2 knights, 2 bishops, 1 king and 1 queen in white. similarly we have 8,2,2,2,1,1 sets of unique pieces in black. Method 1: A = number of ways of selecting 5 out of the above B = number of ways of selecting 5 such that no pawn is selected C = number of ways of selecting 5 such that only 1 or 0 white pawn is selected then Answer = (A-C)/(A-B) A = coeff of x^{5} in (1-x^{9})^{2}(1-x^{3})^{6}(1+x)^{4}(1-x)^{-8 }= 2540 B = coeff of x^{5} in (1-x^{3})^{6}(1+x)^{4}(1-x)^{-6 }= 876 C = coeff of x^{5} in (1+x)(1-x^{9})(1-x^{3})^{6}(1+x)^{4}(1-x)^{-7}=2230 Answer = 310/1664 = 0.186298 Method 2 A = no. of ways of selecting 5 such that 2 or 3 or 4 or 5 white pawns are selected B = no. of ways of selecting 5 C = no. of ways of selecting 5 without any pawns Answer = A/(B-C) B = (all 5 pieces are similar) + (4 similar and 1 different) + (3 sim, 2 sim) + (3 sim,2 different) + (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff) = (^{2}C_{1}) + (^{2}C_{1}*^{11}C_{1}) + (^{2}C_{1}*^{7}C_{1}) + (^{2}C_{1}*^{11}C_{2}) + (^{8}C_{2}*10) + (^{8}C_{1}*^{11}C_{3}) + (^{12}C_{5}) = 2540 C = (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff) = (^{6}C_{2}*^{8}C_{1}) + (^{6}C_{1}*^{9}C_{3}) + (^{10}C_{5) }= 876 A = (5 white pawn) + (4 WP, 1 different) + (3 WP, 2 similar pieces) + (3 WP, 2 different) + (2 WP, 2 sim,1 diff) + (2 WP, 3 diff) = 1 + (^{11}C_{1}) + (^{7}C_{1} + ^{11}C_{2}) + (^{7}C_{1}*^{10}C_{1}) + (^{11}C_{3}) = 310 Answer = 310/1664 = 0.186298 0 Share this post Link to post Share on other sites

0 Posted 21 Apr 2014 · Report post there are 8 pawns, 2 rooks, 2 knights, 2 bishops, 1 king and 1 queen in white. similarly we have 8,2,2,2,1,1 sets of unique pieces in black. Method 1: A = number of ways of selecting 5 out of the above B = number of ways of selecting 5 such that no pawn is selected C = number of ways of selecting 5 such that only 1 or 0 white pawn is selected then Answer = (A-C)/(A-B) A = coeff of x^{5} in (1-x^{9})^{2}(1-x^{3})^{6}(1+x)^{4}(1-x)^{-8 }= 2540 B = coeff of x^{5} in (1-x^{3})^{6}(1+x)^{4}(1-x)^{-6 }= 876 C = coeff of x^{5} in (1+x)(1-x^{9})(1-x^{3})^{6}(1+x)^{4}(1-x)^{-7}=2230 Answer = 310/1664 = 0.186298 Method 2 A = no. of ways of selecting 5 such that 2 or 3 or 4 or 5 white pawns are selected B = no. of ways of selecting 5 C = no. of ways of selecting 5 without any pawns Answer = A/(B-C) B = (all 5 pieces are similar) + (4 similar and 1 different) + (3 sim, 2 sim) + (3 sim,2 different) + (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff) = (^{2}C_{1}) + (^{2}C_{1}*^{11}C_{1}) + (^{2}C_{1}*^{7}C_{1}) + (^{2}C_{1}*^{11}C_{2}) + (^{8}C_{2}*10) + (^{8}C_{1}*^{11}C_{3}) + (^{12}C_{5}) = 2540 C = (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff) = (^{6}C_{2}*^{8}C_{1}) + (^{6}C_{1}*^{9}C_{3}) + (^{10}C_{5) }= 876 A = (5 white pawn) + (4 WP, 1 different) + (3 WP, 2 similar pieces) + (3 WP, 2 different) + (2 WP, 2 sim,1 diff) + (2 WP, 3 diff) = 1 + (^{11}C_{1}) + (^{7}C_{1} + ^{11}C_{2}) + (^{7}C_{1}*^{10}C_{1}) + (^{11}C_{3}) = 310 Answer = 310/1664 = 0.186298 Really sorry for showing the approach here, but this was my first post immediately after joining and I didnt know I could hide spoilers or how to hide them. 0 Share this post Link to post Share on other sites

0 Posted 22 Apr 2014 · Report post no biggie, in the future, just type [ spoiler ]message[ /spoiler ]but without the spaces. 0 Share this post Link to post Share on other sites

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Five pieces are randomly selected from a standard chess set. If it is known that at least one of these pieces is a pawn, what is the probability that at least two of these five pieces are white pawns?

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