Find the probability that he will get atleast one book published if he writes two.

Question

An author writes a good book with a probability of $\frac{1}{2}$.If it is good it is published with a probability of $\frac{2}{3}.$if it is not,it is published with a probability of $\frac{1}{4}$.Find the probability that he will get atleast one book published if he writes two.

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The probability that he will get atleast one book published if he writes two,is $P(x=1)+P(x=2)$,where $P(x=1)$ is the probability that his one book publishes and $P(x=2)$ is the probability that his two books publish.

I found $P(x=1)$ and $P(x=2)$ by using conditional probability.

Probability that his book is good is $\frac{1}{2}$ and Probability that his book is not good is $\frac{1}{2}$.

$P(x=1)=\frac{1}{2}\times\frac{2}{3}+\frac{1}{2}\times\frac{1}{4}=\frac{11}{24}$ $P(x=2)=\frac{11}{24}\times\frac{11}{24}$

The probability that he will get atleast one book published if he writes two,is $P(x=1)+P(x=2)=\frac{11}{24}+\frac{11}{24}\times\frac{11}{24}=\frac{11(24+11)}{24\times24}=\frac{385}{576}$

But the answer given is $\frac{407}{576}$.I dont know where i have gone wrong?

Answer 1

Your calculation of $P(X=1)$ is wrong. Two cases must be looked at: i) the first book is published and the second is not published ii) the first book is not published and the second is published.

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More efficient route:

The probability for one book to be published is $\frac{1}{2}\frac{2}{3}+\frac{1}{2}\frac{1}{4}=\frac{11}{24}$.

So the probability that two books are both not published is $\left(1-\frac{11}{24}\right)\left(1-\frac{11}{24}\right)=\frac{169}{576}$.

Then the probability that at least one book is published is $1-\frac{169}{576}=\frac{407}{576}$

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If $X$ denotes the number of books that will be published, then:

• $P(X=0)=\frac{13}{24}\frac{13}{24}$
• $P(X=1)=\frac{11}{24}\frac{13}{24}+\frac{13}{24}\frac{11}{24}$
• $P(X=2)=\frac{11}{24}\frac{11}{24}$

Binomial distribution with parameters $n=2$ and $p=\frac{11}{24}$.

Answer 2

Let us stay with your line of reasoning and see if you can correct it.

Let $G_i$ be the event that book $i$ is good and let $B_i$ be the event that book $i$ is bad. This gives the possible outcomes $\{(G_1,G_2),(G_1,B_2),(B_1,G_2),(B_1,B_2)\}$. To determine the probability mass function of the number of books published, $X$, you should indeed condition on the four possible outcomes. Thus,

\begin{align} \mathbb{P}(X = 1) &= \mathbb{P}(X = 1 \mid (G_1,G_2)) \cdot \mathbb{P}((G_1,G_2)) \\ &\quad + \mathbb{P}(X = 1 \mid (G_1,B_2)) \cdot \mathbb{P}((G_1,B_2)) \\ &\quad + \mathbb{P}(X = 1 \mid (B_1,G_2)) \cdot \mathbb{P}((B_1,G_2)) \\ &\quad + \mathbb{P}(X = 1 \mid (B_1,B_2)) \cdot \mathbb{P}((B_1,B_2)). \end{align}

As a further hint

\begin{align} \mathbb{P}(X = 1 \mid (G_1,G_2)) &= \mathbb{P}(\text{Book 1 is published and book 2 is not} \mid (G_1,G_2)) \\ &\quad + \mathbb{P}(\text{Book 2 is published and book 1 is not} \mid (G_1,G_2)). \end{align}

Can you take it from here?

Answer 3

You shouldn't be adding P(x=1) and P(x=2). What you need is 1 - (!P(x=1))( !P(x=1)), which is one minus the probability that neither of 2 books get published. This is the probability that at least one of the 2 books does get published.

Think about it this way: if I write N books, each with a probability of success P, what's the chance that NONE of those books get published? Anything other than that (1 minus that chance), and I've succeeded in publishing at least one book.

Answer 4

1. What is the probability $p$ that he gets published? He either writes a good book and gets published or he writes a bad one and gets published. Thus $$p = \frac{1}{2}\cdot\frac{2}{3}+\frac{1}{2}\cdot\frac{1}{4} = \frac{11}{24}. $$

2. What is that probability that he gets published if he writes two books? Well, assuming each book is independent of another, then there are $n = 2$ independent trials (books) with probability $p= \frac{11}{24}$ of success (getting published). Thus, if $X$ is number of books published, then $X$ follows a binomial distribution with parameters $n,p$. So, $$P(X\geq 1) = \binom{2}{1}p^1(1-p)^1+\binom{2}{2}p^2(1-p)^0 = \frac{407}{576}.\tag{$\star$}$$ You're missing a factor of $2$ in your calculation. Remember, when the author gets published once, there are two choices. Either the author gets published in the first trial or in the second. Thus there are $2$ choices. Compare your calculation with $(\star)$.