Question

Two dices thrown simultaneously find probability on which-on 1st dice prime no. and on 2nd dices divisible by 2 no. will occurs​

Answer 1

$\LARGE\textbf{\underline{\underline{✪\: \: \: Solution :-}}}$

• The Sample Space ( S ) when two dices are thrown simultaneously :-

➛ S = {( 1 , 1 ) , ( 1 , 2 ) , ( 1, 3 ) , ( 1, 4 ) , ( 1 , 5 ) , ( 1 , 6 ) ,( 2 , 1 ) , ( 2 , 2 ) , ( 2 , 3 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 2 , 6 ) , ( 3 , 1 ) , ( 3 , 2 ) , ( 3 , 3 ) , ( 3 ,4 ) , (3 , 4 ) , ( 3 , 5 ) , ( 3 , 6 ) , ( 4 , 1 ) , ( 4 ,2 ) , ( 4 , 3 ) , ( 4 ,4 ) , ( 4 ,5 ) , ( 4 , 6 ) , ( 5 , 1 ) , ( 5 , 2 ) , (.5 , 3 ) , ( 5 ,4 ) , ( 5 , 5 ) , ( 5 , 6 ) , ( 6 , 1 ) , ( 6 , 2 ) , ( 6 , 3 ) , ( 6 , 4 ) , ( 6 , 5 ) , ( 6 , 6 ) }

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➛ n ( S ) = 36 .

• Let A be the event of getting ' Prime no. ' on 1ˢᵗ dice and ɡettinɡ a ' Number divisible by 2 ' on 2ⁿᵈ :-

➛ ( A ) = { ( 2 , 2 ) , ( 2 , 4 ) , ( 2 , 6 ) , ( 3 , 2 ) , ( 3 , 4 ) , ( 3 , 6 ) , ( 5 , 2 ) , ( 5 , 4 ) , ( 5 , 6 ) } .

➛ n ( A ) = 9 .

$\large:\: \bigstar\textsf\textcolor{orange}{\: \: \: P ( A ) = \cfrac{ n ( A ) }{ n ( S ) } }\\\\\\\large: \: \Longrightarrow\textsf{= \cfrac{9 }{36 } }\\\\\\\large: \: \Longrightarrow\textsf{= \cfrac{ \cancel 9 }{ \cancel 36} }\\\\\\\large : \: \Longrightarrow\underline{\boxed{\textsf\textcolor{purple}{P ( A ) = \cfrac{1}{4} }}}$