Two dice are thrown simulta-neously. Find the probability of getting:

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asked Apr 6 in Class X Maths by muskan15 (-3,352 points)

Two dice are thrown simulta-neously. Find the probability of getting:
(i) an even number as the stun, (ii) the sum as a prime number, (iii) a total of at least 10, (iv) a doublet of even number, (v) a multiple of 3 as the sum.

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1 Answer

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answered Apr 6 by navnit40 (-4,342 points)

Solution: n(S) = 36

Such as

(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), (2), (3, 3), (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).

(i) Event = {even number as the sum}

I,e., = {(1, 1), (1, 3), (3, 1), (2, 2), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (2, 6), (6, 2), (4, 4), (5, 3), (3, 5),( 5, 5), (6, 4), (4, 6) and (6,6)}

n(E) = 18

P (E) = ?

∴ P (E) = n (E)/n (S) = 18/36 = ½.

(ii) Event = {the sum as a prime number}

i.e., = {(1, 1), (1, 2), (2, 1), (1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5) and (5, 6)}

n(E) = 15

P (E) = ?

∴ P (E) = n (E)/n (S) = 15/36 = 5/12.

(iii) Event = {getting a total of at least 10}

= {(6, 4), (4, 6), (5, 5), (6, 5), (6, 6)}

n(E) = 6

P (E) =?

∴ P (E) = n (E)/n (S) = 6/36 = 1/6.

(iv) Event = {getting a doublet of even number}

= {(2, 2), (4, 4), (6, 6)}

n(E) = 3

P (E) =?

∴ P (E) = n (E)/n (S) = 3/36 = 1/12.

(v) Event = {multiple of 3 as a sum}

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

n(E) = 12

P (E) =?

∴ P (E) = n (E)/n (S) = 12/36 = 1/3.

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