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India plays two matches each with West Indies and Australia. In any match, the probabilities of India getting, points 0, 1 and 2 are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is 

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Since there are 4 matches to be played.India can get a maximum of 8 points(2 points in each match).

Probability of getting exactly 7 points = P(Getting 2 points each of 3 matches and getting 1 point in one match)

=$\binom{4}{3}$ (0.5)3 (0.05)1 =0.025

Probability of getting exactly 8 points =P(Getting 2 points in each of 4 matches)

=$\binom{4}{4}$ (0.5)4 (0.05)0=0.0625

P(india gets at least 7 points) =P(Getting exactly 7 points) + P(Probability of getting Exactly 8 points)

                                            = 0.025 + 0.0625

                                            =0.0875

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Let if India loose gets point 0 Draws gets point 1 Wins gets point 2 
India will get point 7 if it wins 3 matches and Draws 1 
India will get point 8 if it wins all the 4 matches

Therefore combination of winning 3 matches 2 win with westindies and 1 win and 1 draw with australia  or vice versa 

.5*.5*.5*.05*4 = .025

Getting point 2 all win .5*.5*.5*.5 = .0625 

P(x>=7) = .025+ .0625 = .875

 

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