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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

dato datuashvili
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2 Answers2

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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

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

Brian M. Scott
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  • they are independent and i have multiplied their probability,that why i got $1/8$ – dato datuashvili Aug 01 '13 at 07:23
  • @dato: What you overlooked is that the probability of $2$ evens and an odd is actually $3\cdot\frac18$, because the odd can be the first, second, or third roll. Each of the three possibilities occurs with probability $\frac18$, so the total probability of getting $2$ evens and an odd is $\frac38$. – Brian M. Scott Aug 01 '13 at 07:27
  • for example probability of odd number in each time is $1/2$,so i have multiplied on each other and i have got $1/2$$1/2$$1/2$ – dato datuashvili Aug 01 '13 at 07:27
  • @dato: Yes, $\frac18$ is correct for the probability of each of the $8$ possible outcomes. Using E for even and O for odd, these outcomes are OOO, OOE, OEO, EOO, OEE, EOE, EEO, and EEE. The last $4$ in that list have at least two successes. Each has probability $\frac18$, so the total probability of those $4$ outcomes is $4\cdot\frac18=\frac12$. – Brian M. Scott Aug 01 '13 at 07:29
  • for two event it is easy to count it by hand,but for the three event which formula i should use for save time?because on GRE time is very important you know about it :) – dato datuashvili Aug 01 '13 at 07:30
  • @dato: In a problem that asks for the probability of $\ge k$ successes, when the maximum possible number is $m$, I usually start by asking whether $k$ is closer to $m$ or closer to $0$. If it’s closer to $m$, the fastest solution may be to calculate it directly, by adding the probabilities of $k$ successes, $k+1$ successes, ..., $m$ successes. If it’s closer to $0$, I’ll calculate the probabilities of $0,1,\ldots,k-1$ successes, add those, and subtract from $1$. Unless I see a really fast solution quickly, it’s usually quicker to use whichever of these straightforward approaches is shorter. – Brian M. Scott Aug 01 '13 at 07:36
  • in this case it is easy to count by hand?when numbers are small – dato datuashvili Aug 01 '13 at 07:39
  • @dato: Yes. When the number of trials ($m$ in my previous comment) is large, the calculation can get quite messy unless $k$ is very small or very close to $m$. You might want to take a look at this. – Brian M. Scott Aug 01 '13 at 07:42
  • in this case number of trials would be $3$ right?and we would have combination elements $2$ from $3$ and $3$ from $3$ and sum of this right?each probability $1/2$,because odd and even numbers have same probability $1/2$ – dato datuashvili Aug 01 '13 at 07:50
  • @dato: Yes, this experiment consists of $3$ trials. Thus, it can result in $0,1,2$, or $3$ successes. In general if there are $m$ trials, you can get $0,1,2,\ldots,m$ successes. – Brian M. Scott Aug 01 '13 at 07:51
  • i think i am checking my week points and correcting it – dato datuashvili Aug 01 '13 at 07:54
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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 $1/8+3/8=1/2$.

Daniel Montealegre
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