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# two pawns

## Question

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?

## 5 answers to this question

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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

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

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

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

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is the answer 3874 / (32c5) ??

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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.

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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

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

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

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

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.

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

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