Question

Need help on probability. Please explain how you did it so I can know how to do it.

Answer 1

Answer:

$\text{P}(B \text{ or }C)=\dfrac{5}{6}$

Step-by-step explanation:

Mutually Exclusive Events

For two events, A and B, where A and B are mutually exclusive:

$\boxed{\text{P}(A \text{ or }B)=\text{P}(A)+\text{P}(B)}$

Probability distribution table:

$\begin{array}{|c|c|c|c|c|c|c|c|}\cline{1-7} x & 1 & 2 & 3 & 4 & 5 & 6 \\\cline{1-7} \text{P}(X=x)\phantom{\dfrac11}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\\cline{1-7}\end{array}$

where X is the score on a fair, six-sided dice.

P(B or C) means "the probability of rolling an even number or a multiple of 3".

As an even number of a fair, six-sided dice can never be a multiple of 3, the two events B and C are mutually exclusive.

Calculate the probabilities for events B and C.

Event B

Rolling an even number.

$\begin{aligned}\implies \text{P}(X \text{ is even})&=\text{P}(2 \text{ or }4\text{ or }6)\\\\&=\text{P}(X=2)+\text{P}(X=4)+\text{P}(X=6)\\\\& = \dfrac{1}{6}+ \dfrac{1}{6}+ \dfrac{1}{6}\\\\&=\dfrac{3}{6}\end{aligned}$

Event C

Rolling a multiple of 3.

$\begin{aligned}\implies \text{P}(X \text{ is multiple of 3})&=\text{P}(3 \text{ or }6)\\\\&=\text{P}(X=3)+\text{P}(X=6)\\\\& = \dfrac{1}{6}+ \dfrac{1}{6}\\\\&=\dfrac{2}{6}\end{aligned}$

Solution

$\begin{aligned}\implies \text{P}(B \text{ or }C)&=\text{P}(B)+\text{P}(C)\\\\& = \dfrac{3}{6}+\dfrac{2}{6}\\\\& = \dfrac{5}{6}\end{aligned}$

Answer 2

Answer:

I believe it should be 5/6

Step-by-step explanation:

Here's why, the probability of getting a B, or an even number is 3/6 since there are 3 even numbers on a dice. The probability of getting a multiple of 3 is 2/6 since the only multiples of 3 are 1 and 3. Add the 2 together to get 5/6. Hope this helps.