[Math] the probability that at least $8$ teenagers in a group of $10$ watched a rented movie at least once last week

binomial distributionprobabilityprobability distributions

The probability that a randomly selected teenager watched a rented video at least once during a week was $0.73$. What is the probability that at least $8$ teenagers in a group of $10$ watched a rented movie at least once last week? (Round your answer to four decimal places.)

$P(X \geq 8) =$

So for this one I did

$1-{_{10}\mathsf C}_8(.73)^9(.27)^1=.2847102215$ and that rounds to $.2487$

But that is not the correct answer, can someone tell me where I went wrong and how to fix it
I also did this question with exponents as an $8$ and a $2$ but this was also wrong

Best Answer

It's not clear what your answer actually calculates.

I assume you meant $$\binom{10}{\color{red}{9}}(0.73)^9(.27)^{10-9}$$ which is actually equal to $P(X=9)$. So under that assumption, what you have calculated is $$1-\binom{10}{9}(0.73)^9(.27)^{10-9} = 1-P(X=9) = P(X\neq 9).$$ This is calculating the chance that $X$ is anything but $9$.

What you want is $$P(X\geq 8) = P(X = 8\cup X=9 \cup X=10) = P(X=8)+P(X=9)+P(X=10)$$ where you can just sum up the values since the events are disjoint. Similarly, if you want to use the complement, then $$P(X\geq 8) = 1-P(X<8) = 1-\sum_{k=0}^{7}P(X=k).$$

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[Math] Binomial Distribution Problem Involving Excel

The correct answer is equivalent to not having all ten not O-negative, i.e. $1-(1-0.07)^{10} \approx 0.516$. Any of the following Excel formulae will give this:

=1-(1-0.07)^10

=1-BINOM.DIST(0,10,0.07,TRUE)

=1-BINOM.DIST(0,10,0.07,FALSE)

=1-BINOM.DIST(10,10,0.93,FALSE)

=BINOM.DIST(9,10,0.93,TRUE)

[Math] Probability answer cross check

Your calculation cannot be right. For any probability must be between $0$ and $1$, but $3.2$ is not.

We have repeated an experiment independently $16$ times. Each time the probability of success (prefers red) is $0.4$. We want the probability that the number of successes is $\ge 2$.

The probability that the number of successes is $k$ is $\binom{16}{k}(0.4)^k(0.6)^{16-k}$. For the probability that the number of successes is $\ge 2$, add up, $k=2$ to $16$. This will give you a correct expression, and with a great deal of work you can evaluate the $15$ terms and find their sum.

However, there is a simpler way. Let $p$ be the probability of $0$ or $1$ successes. Then the probability of $\ge 2$ successes is $1-p$. If we can find $p$, we can then easily compute the answer.

The probability of $0$ successes (nobody in our sample prefers red) is $(0.6)^{16}$.

The probability exactly one person prefers red is $\binom{16}{1}(0.4)(0.6)^{15}$.

So $p=(0.6)^{16}+\binom{16}{1}(0.4)(0.6)^{15}$. Calculate, and to finish find $1-p$.

Best Answer

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[Math] Probability answer cross check

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