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Real Numbers

Question
CBSEENMA10009375
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Two dice are thrown simultaneously. Find the probability of getting :
(i) a sum less than 6 (ii) a sum less than 7
(iii) a sum more than 7 (iv) 8 as the sum

Solution

If two dice are thrown simultaneously, then possible outcomes(s) are 36. i.e., n(S)= 36
(i) Let E be the favourable outcomes of getting a sum less than 6, then
E= {(1,1),(1, 2), (l, 3), (1,4), (2, 1), (2, 2), (2, 3), (3, l), (3, 2), (4, l)}
⇒ n(E)= 10 Therefore,
P(E) = fraction numerator straight n left parenthesis straight E right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 10 over 36 equals 5 over 18

(ii) Let F be the favourable outcomes of getting a sum less than 7.
F= {(1, 1), (1, 2), (1,3), (1,4), (1,5). (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4,1), (4, 2), (5, 1)}
⇒ n( F)=15 Therefore,
P(F) = fraction numerator straight n left parenthesis straight F right parenthesis space over denominator straight n left parenthesis straight S right parenthesis end fraction equals 15 over 36 equals 5 over 12

(iii) Let G be the favourable outcomes of getting a sum more than 7, then
G = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)1
⇒ n(G) = 15 Therefore,
P(G) = fraction numerator straight n left parenthesis straight G right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 15 over 36 equals 5 over 12

(iv) Let H be the favourable outcomes of getting 8 as the sum. then
H = |(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
⇒ n( H) = 6 Therefore.
P(H) = fraction numerator n left parenthesis H right parenthesis over denominator n left parenthesis S right parenthesis end fraction equals 5 over 36

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