A pair of dice is thrown once. find the probability of getting the same number on each dice]
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Heya User,
Consider the Events:-->
-->Sample Space = [ 6 ]³ = 216 <--- Soo, 216 possible outcomes....
Now,
--> How can we write 10...
-----> 10 = a + b + c ∀ 0<(a,b,c) < 7
So, all such possible outcomes are :->
--> When a = 1 -->
--->> 10 = 1 + 3 + 6 ; 1 + 6 + 3
--->> 10 = 1 + 4 + 5 ; 1 + 5 + 4
When a = 2;
--->> 10 = 2 + 2 + 6; 2 + 6 + 2
--->> 10 = 2 + 3 + 5; 2 + 5 + 2
--->> 10 = 2 + 4 + 4; and 2 + 4 + 4 where (b,c) are doublets....
Similarly, for a = 3, there are 6 outcomes, for a = 4, 8 outcomes,
---> for a = 5, 4 outcomes, and for a = 6, 4 outcomes
Hence, n(E) = 4 + 6 + 6 + 8 + 4 + 4 = 32
Hence, P(E) = 32/216 = 4/27 .... <-- ANS
Consider the Events:-->
-->Sample Space = [ 6 ]³ = 216 <--- Soo, 216 possible outcomes....
Now,
--> How can we write 10...
-----> 10 = a + b + c ∀ 0<(a,b,c) < 7
So, all such possible outcomes are :->
--> When a = 1 -->
--->> 10 = 1 + 3 + 6 ; 1 + 6 + 3
--->> 10 = 1 + 4 + 5 ; 1 + 5 + 4
When a = 2;
--->> 10 = 2 + 2 + 6; 2 + 6 + 2
--->> 10 = 2 + 3 + 5; 2 + 5 + 2
--->> 10 = 2 + 4 + 4; and 2 + 4 + 4 where (b,c) are doublets....
Similarly, for a = 3, there are 6 outcomes, for a = 4, 8 outcomes,
---> for a = 5, 4 outcomes, and for a = 6, 4 outcomes
Hence, n(E) = 4 + 6 + 6 + 8 + 4 + 4 = 32
Hence, P(E) = 32/216 = 4/27 .... <-- ANS
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