A die is thrown twice. What is the probability that at least one of th

1.	A die is thrown twice. What is the probability that at least one of the two throws comes up with the number 4?
Answer» Given : A die is thrown twice To find : Probability that at least one of the two throws comes up with the number 4 The formula used : Probability = \(\frac{favourable\,number\,of\,outcomes}{Total\,no.of\,outcomes}\) A die is numbered from 1 to 6 When a die is thrown twice, total number of outcomes = 6² = 36 Favourable outcomes = {(4,1) ,(4,2) ,(4,3) ,(4,4) ,(4,5) ,(4,6) ,(1,4) ,(2,4) ,(3,4) ,(5,4) , (6,4)} Favourable number of outcomes = 11 Probability that at least one of the two throws comes up with the number 4 = \(\frac{11}{36}\) The probability that at least one of the two throws comes up with the number 4 = \(\frac{11}{36}\)

A die is thrown twice. What is the probability that at least one of the two throws comes up with the number 4?

Answer»

Given : A die is thrown twice

To find : Probability that at least one of the two throws comes up with the number 4

The formula used : Probability

= \(\frac{favourable\,number\,of\,outcomes}{Total\,no.of\,outcomes}\)

A die is numbered from 1 to 6

When a die is thrown twice, total number of outcomes = 6² = 36

Favourable outcomes =

{(4,1) ,(4,2) ,(4,3) ,(4,4) ,(4,5) ,(4,6) ,(1,4) ,(2,4) ,(3,4) ,(5,4) , (6,4)}

Favourable number of outcomes = 11

Probability that at least one of the two throws comes up with the number 4 = \(\frac{11}{36}\)

The probability that at least one of the two throws comes up with the number 4 = \(\frac{11}{36}\)

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