If three square are selected at random from chess board. then the probability that they form the letter 'L' is (a) `196/(64C_3)` (b) `49/(64C_3)` (c) `36/(64C_3)` (d) `98/(64C_3)`

Options

A. `196//.^(64)C_(3)`
B. `49//.^(64)C_(3)`
C. `36//.^(64)C_(3)`
D. `98//.^(64)C_(3)`

Answer

A

Explanation

We have,
n(S) = `.^(64)C_(3)`
Let 'E' be the event of selecting 3 squares which form the letter 'L'.
The number of ways of selecting squares consisting of 4 unit squares is `7 xx 7 = 49`.
Each square with 4 unit squares form 4 L-shapes consisting of 3 unit squares. Therefore.
n(E) = `7 xx 7 xx 4 = 196`
`therefore P(E) = (196)/(.^(64)C_(3))`