probability calculation in dice throwing

Question

i would like to understand what i am doing wrong in the following problem solution,problem is :

A dice is thrown $3$ times. Getting a multiple of $2$ is considered a success. Find the probability of at least two successes.

this problem i have translated as:at least two success means that at three throwing,at least two should be multiply of $2$,or on the other hand this prob-ability equal $1$-neither multiply of $2$ or $1$,if i denote neither multiply of two by $A$ and $1$ multiply as $B$,we have

$1-p(A or B)$;

now let us write formula for $P(A or B)=P(A)+P(B)-P(A and B)$

multiply of $2$ are $A={2,4,6}$ and complement of $A$ is $A'={1,3,5}$ ,in each throwing occurring of not multiply of $2$ is $1/2$,so in total it would be $1/8$,now probability of two non multiply of two and $1$ multiply is $1/8$ again so we have; $1/8+1/8-P(A and B)$,now what about probability of $P(A and B)$?would it be $1/8*1/8$ or $1/4*1/2$?

because probability of two non multiply of

$2$ is $1/4$ and $1/2$ for multiply of $2$? please help me

Answer 1

When you roll a die $3$ times, you can get $0,1,2$, or $3$ successes. The question asks for the probability that you get either $2$ or $3$ successes. The straightforward solution is to calculate the probability $p_2$ of $2$ successes and the probability $p_3$ of $3$ successes and add these two numbers. Since the probability of a success on any given roll is $\frac12$, $p_3=\left(\frac12\right)^3=\frac18$. Calculating $p_2$ is almost as easy: there are $3$ ways to get $2$ successes and a failure (SSF, SFS, FSS), and each has probability $\left(\frac12\right)^3=\frac18$, so $p_2=\frac38$. The probability of at least $2$ successes (i.e., $2$ or $3$ successes), is therefore $p_2+p_3=\frac18+\frac38=\frac12$.

There is also a ‘clever’ way to solve it. Notice that since the probability of a success is the same as the probability of a failure, the probability of getting $2$ or $3$ failures must be the same as the probability of getting $2$ or $3$ successes. But every possible outcome of rolling the die $3$ times is either $\ge 2$ successes or $\ge 2$ failures: if you have $0$ or $1$ success, you have $3$ or $2$ failures. And no outcome gives you both $\ge 2$ successes and $\ge 2$ failures. Thus,

$$\Bbb P(\ge 2\text{ successes})=\Bbb P(\ge 2\text{ failures})$$

and

$$\Bbb P(\ge 2\text{ successes})+\Bbb P(\ge 2\text{ failures})=1\;,$$

so

$$\Bbb P(\ge 2\text{ successes})=\Bbb P(\ge 2\text{ failures})=\frac12\;.$$

Answer 2

First of all I am just going to say "even" instead of multiple of two, and odd if it is not a multiple of two (im doing this because your text was somewhat hard to read). Your mistake is that the probability of having a total of two odds and one even is not $\frac{1}{8}$. The possibilities are: Even-odd-odd, odd-even-odd, even-odd-odd. Each one has probability of $\frac{1}{8}$ so this event has probability $\frac{3}{8}$.

The other mistake is that the probability of $A$ and $B$ both ocurring is zero since you cant have exactly one even and exactly zero evens at the same time. Hence at the end you you: $1-(1/8+3/8)=1/2$

Another way you could calculate it would be directly: Probability of two successes is $3/8$ and probability of one success is $1/8$ the events are disjoint, so their union gives you $1/8+3/8=1/2$.