1.

Counters marked 1, 2, 3 are placed in a bag and one is drawn and replaced. The operation is repeated three times. The chance of obtaining a total of 6 is(a) \(\frac{2}{9}\) (b) \(\frac{7}{27}\)(c) \(\frac{13}{17}\)(d) \(\frac{14}{27}\)

Answer»

(b) \(\frac{7}{27}.\)

Let S be the sample space of drawing a counter three times and replacing it each time. 

Then, n(S) = 3 × 3 × 3 = 27 

Let A : Event of obtaining a total of 6 in the three draws of counters. 

Then, A = {(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1), (2, 2, 2)} 

⇒ n(A) = 7 

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{7}{27}.\)


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