Question

Pls answer! I don’t know math @ all

Answer 1

The probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4 is $\frac{5}{12}$.

Solution:

Total number of outcomes N(S) = 36

Let A be the sum that is a multiple of 6 and

B be the sum that is a multiple of 4.

Sum that is a multiple of 6 = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)

N(A) = 6

Sum that is a multiple of 4 = (1, 3), (2, 2), (2, 6), (3, 1), (3, 5),

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

N(B) = 9

$P(A)=\frac{N(A)}{N(S)}$

$P(A)=\frac{6}{36}=\frac{1}{6}$

$P(B)=\frac{N(B)}{N(S)}$

$P(B)=\frac{9}{36}=\frac{1}{4}$

Probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4:

$P(A \cup B)=P(A)+P(B)$

$P(A \cup B)=\frac{1}{6} +\frac{1}{4}$

               $=\frac{2+3}{12}$ (Make the denominator same)

               $=\frac{5}{12}$

Hence the probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4 is $\frac{5}{12}$.