Question

find the probability of a four turning up at least once in two tosses of a fair die

Answer 1

total outcomes -36
fav outcomes, _(4,4),(1,4),(2,4),(3,4),(5,4,),(6,4),(4,1),(4,2),(4,3),(4,5),(4,6)
p(atleast 1 four)- 11/36

Answer 2

The probability of a four turning up at least once in two tosses of a fair die is $\frac{11}{36}$

Solution:

Sample space when two tosses of a fair die is:

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

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

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

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

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

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

Total number of possible outcomes = 36

Favorable outcome = four turning up at least once in two tosses of a fair die

Favorable outcome = (1, 4) , (2, 4) , (3, 4) , (4,1) ,(4,2), (4,3), (4,4) ,(4,5), (4,6)  , (5, 4) , (6, 4)

Number of favorable outcome = 11

The probability of an event is given as:

$Probability = \frac{ \text{ number of favorable outcomes }}{ \text{ total number of possible outcomes }}\\\\Probability = \frac{ 11}{36}$

Thus, the probability of a four turning up at least once in two tosses of a fair die is $\frac{11}{36}$

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