If value of die is 1 2 or 3 then die is rolled a second time. Probability that sum of values will be at least 6
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Here our sample space consists of 3+3×6=21 events- (4),(5),(6),(1,1),(1,2)…(3,6).(4),(5),(6),(1,1),(1,2)…(3,6).
Favorable cases =(6),(1,5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6).
Required Probability
=No. of favorable cases/Total cases =10/21
But this is wrong way of doing. Because due to 2 tosses for some and 1 for some, individual probabilities are not the same. i.e., while (6)(6) has 1/6probability of occurrence, (1,5)has only 1/36 probability. So, our required probability
⇒1/6+(9×1/36)=5/12.
Favorable cases =(6),(1,5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6).
Required Probability
=No. of favorable cases/Total cases =10/21
But this is wrong way of doing. Because due to 2 tosses for some and 1 for some, individual probabilities are not the same. i.e., while (6)(6) has 1/6probability of occurrence, (1,5)has only 1/36 probability. So, our required probability
⇒1/6+(9×1/36)=5/12.
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