for 6. require you to buy 10 cards to finally win a prize; and Platinum cards re
ID: 3371434 • Letter: F
Question
for 6. require you to buy 10 cards to finally win a prize; and Platinum cards require you on average to buy 12 cards to finally win a prize. Each customer consistently buys only one kind of card 30% of customers prefer Gold cards, and 70% prefer Platinum cards. A convenience store sells two kinds of lottery cards: Gold cards on average customer will finally a) (5 points) What is the probability a randomly chosen have won before buying the 8th card? b) (5 points) A customer comes out of the store and announces "I just won my first prize, but it was the 15th card I bought!" What is the probability that customer buys Platinum cards?Explanation / Answer
a) In case of gold card, probability that the customer wins the prize before the 8th card is computed as:
= 1 - (1 - 0.1)7 = 0.5217
Now for the platinum card the same probability is computed as:
= 1 - (1 - (1/12) )7
= 0.4561
Therefore now probability of a randomly chosen customer that would have finally won before buying the 8th card is computed as:
= 0.3*(0.5217) + 0.7*(0.4561) = 0.4758
Therefore 0.4758 is the required probability here.
b) Probability that the person wins on 15th card given that he buys a gold card is computed as:
= 0.914*0.1 = 0.0229
Probability that the person wins on 15th card given that he buys the platinum card is computed as:
= (11/12)14*(1/12)
= 0.0246
Therefore probability that he wins on 15th day is now computed using the law of total probability as:
= 0.0229*0.3 + 0.0246 *0.7
= 0.0241
Therefore now given that he won on the 15th card, probability that the customer had bought a platinum card is computed as:
= (0.0246 *0.7) / 0.0241
= 0.7138
Therefore 0.7138 is the required probability here.
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