A bite the dust is tossed twice. What is the likelihood that
A bite the dust is tossed twice. What is the likelihood that

(I) 5 won’t come up one or the other time?

(ii) 5 will come up in some measure once?

[Hint : Throwing a bite the dust twice and tossing two dice at the same time are treated as the equivalent experiment]

Solution:
Results are:

(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, 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)

Along these lines, the complete number of result = 6×6 = 36

(I) Method 1:

A = 5 comes in first toss,

B = 5 comes in second toss

P(A) = 6/36,

P(B) = 6/36 and

P(not B) = 5/6

In this way, P(not A) = 1-(6/36) = 5/6

∴ The necessary likelihood = (5/6)×(5/6) = 25/36

Technique 2:

Leave E alone the occasion in which 5 doesn’t come up one or the other time.

Along these lines, the positive results are [36–(5+6)] = 25

∴ P(E) = 25/36

(ii) Number of occasions when 5 comes once = 11(5+6)

∴ The necessary likelihood = 11/36