two pawns

[BMAD]

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?

[toletyrajesh]

is the answer 3874 / (32c5) ??

[Rainman]

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*40/32⁵-24*23*22*21*20/32⁵ = 1-24*23*22*21*60/32⁵ = 1-23*11*7*5*3³/2¹⁹ = 285203/524288

P(B) = 1-P(no pawns selected) = 1-16*15*14*13*12/32⁵ = 1-13*7*5*3²/2¹⁸ = 258049/262144

P(A)/P(B) = 285203/516098, which would be the probability we seek.

[m00li]

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 x⁵ in (1-x⁹)²(1-x³)⁶(1+x)⁴(1-x)^-8 = 2540

B = coeff of x⁵ in (1-x³)⁶(1+x)⁴(1-x)^-6 = 876

C = coeff of x⁵ in (1+x)(1-x⁹)(1-x³)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)

= (²C₁) + (²C₁*¹¹C₁) + (²C₁*⁷C₁) + (²C₁*¹¹C₂) + (⁸C₂*10) + (⁸C₁*¹¹C₃) + (¹²C₅) = 2540

C = (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff)

= (⁶C₂*⁸C₁) + (⁶C₁*⁹C₃) + (¹⁰C₅₎ = 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 + (¹¹C₁) + (⁷C₁ + ¹¹C₂) + (⁷C₁*¹⁰C₁) + (¹¹C₃) = 310

Answer = 310/1664 = 0.186298

[m00li]

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 x⁵ in (1-x⁹)²(1-x³)⁶(1+x)⁴(1-x)^-8 = 2540

B = coeff of x⁵ in (1-x³)⁶(1+x)⁴(1-x)^-6 = 876

C = coeff of x⁵ in (1+x)(1-x⁹)(1-x³)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)

= (²C₁) + (²C₁*¹¹C₁) + (²C₁*⁷C₁) + (²C₁*¹¹C₂) + (⁸C₂*10) + (⁸C₁*¹¹C₃) + (¹²C₅) = 2540

C = (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff)

= (⁶C₂*⁸C₁) + (⁶C₁*⁹C₃) + (¹⁰C₅₎ = 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 + (¹¹C₁) + (⁷C₁ + ¹¹C₂) + (⁷C₁*¹⁰C₁) + (¹¹C₃) = 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.

[phil1882]

no biggie, in the future, just type [ spoiler ]message[ /spoiler ]but without the spaces.