1708For (2) part of (1) I am Getting \mathbf{=\frac{7}{44}}$
1)Two squares are chosen at random from the 64 squares of a chessboard The probability that they have a side in common is ___
Probability that they have just one vertex in common is ____
2)If P(AUB)=P(A∩B),
(a)P(A)=P(B)
(b)P(AUB)=1
(c)P(A∩B)=1
(d)P(A)+P(B)=1
(e)none of these
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8 Answers
1708\hspace{-16}\mathbf{(1)::}$Choosing $\mathbf{2}$ Square from total $\mathbf{64}$ Square is $\mathbf{\binom{64}{2}=2012}$\\\\ Now taking $\mathbf{2}$ square Horozontailly(Common Side) like $\mathbf{\square\!\square}$\\\\ Then There are $\mathbf{7}$ Possiabilities(Bcz Row has $\mathbf{8}$ boxes in Chess)\\\\ and There are $\mathbf{8}$ rows. in Chess Board\\\\ So Total no. of $\mathbf{2}$ Square each side Common is $\mathbf{=7\times 8=56}$\\\\\\
\bold{Similarly \;\; now\;\; taking\; 2 \; Square \; Vertically \; like} \begin{array}{c}\square \\ [-2mm] \square \end{array}$\\\\ Then There are $\mathbf{7}$ Possiabilities(Bcz Row has $\mathbf{8}$ boxes in Chess)\\\\ and There are $\mathbf{8}$ rows. in Chess Board\\\\ So Total no. of $\mathbf{2}$ Square each side Common is $\mathbf{=7\times 8=56}$\\\\\\ So Total favourable cases is $\mathbf{=56+56=108}$\\\\ So Required Probability $\mathbf{=\frac{108}{2012}=\frac{1}{18}}$
1057simply....
we know P(AUB)=P(A)+P(B) - P(A∩B)
and also P(A∩B) =P(A).P(B)
ATP...
P(A)+P(B)=2P(A).P(B)
or (√P(A)-√P(B))2 =0
therefore P(A)=P(B)
1708Yes I am also getting 7/144
(forget to write 1 in extreme left)
1for 1) 1st part i am also getting 118
for the 2nd part i am also getting 7144