Question

If three one by one squares are selected at random from the chessboard, then the probability that they form the letter ${}^{\prime }{L}^{\prime }$ is

• A. $\frac{196}{{{}^{64}}_{C_3}}$
• B. $\frac{49}{{{}^{64}}_{C_3}}$
• C. $\frac{36}{{{}^{64}}_{C_3}}$
• D. $\frac{98}{{{}^{64}}_{C_3}}$

Answer 1

The correct option is A $\frac{196}{{{}^{64}}_{C_3}}$

Probability is given by,

$P\left(E\right)$$=\frac{\text{no. of possible outcomes}}{\text{No. of total outcomes}}=\frac{n\left(E\right)}{n\left(S\right)}$

$n\left(S\right)={{}^{64}}_{C_3}$
Let ${}^{\prime }{E}^{\prime }$ be the even of selecting $3$ squares which form the letter ${}^{\prime }{L}^{\prime }$.
No. ways of selecting squares consisting of $4$ unit squares $={{}^7}_{C_1}\times {{}^7}_{C_1}=49$
Each square with $4$ unit squares forms $4L-$ shapes consisting of $3$ unit squares.
$\therefore n\left(E\right)=7\times 7\times 4=196$

$\therefore P\left(E\right)=\frac{196}{{{}^{64}}_{C_3}}$