1.

Suppose the probability for A to win game B 0.4. If A has an option of playing either a" best of 3 game" of a "best of 5 game" match against B, which option should be choose so that the probability of his winning the match is higher?

Answer»

The probability p1 (say) of winning the best of three games is = the prob. of winning two games + the prob. of winning three games. 

= 3C2 (0.6) (0.4)2 + 3C3 (0.4)3 

 [Using Binomial distribution] 

Similarly the probability of winning the best five games is p2 (say) = the prob. of winning three games + the prob. of winning 5 games.

 = 5C3 (0.6)2 (0.4)3 + 5C3 (0.6) (0.4)3 + 5C5 (0.4)

We have p1 = 0.288 + 0.064 = 0.352 

And p2 = 0.2304 + 0.0768 + 0.01024 = 0.31744 

As p1 > p2 

∴ A must choose the first offer i.e. best of three games. 



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