Question

Math

Posted 4 months ago

```
Cora is playing a game that involves flipping three coins at once.
Let the random variable $H$ be the number of coins that land showing "heads". Here is the probability distribution for $H$ :
\begin{tabular}{lllll}
$H=$ \# of heads & 0 & 1 & 2 & 3 \\
\hline$P(H)$ & 0.125 & 0.375 & 0.375 & 0.125
\end{tabular}
What is the probability that Cora flips no more than 1 head?
$P($ no more than 1 head $)=$ $\square$
```

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Answer from Sia

Posted 4 months ago

Solution by Steps

step 1

Calculate the probability of flipping no more than 1 head by adding the probabilities of flipping 0 heads and 1 head

step 2

Use the given probabilities from the table: $P(H=0) = 0.125$ and $P(H=1) = 0.375$

step 3

Add the probabilities: $P(H=0) + P(H=1) = 0.125 + 0.375$

step 4

Perform the addition: $0.125 + 0.375 = 0.5$

step 5

Convert the decimal to a fraction: $0.5 = \frac{1}{2}$

Answer

$P($ no more than 1 head $) = \frac{1}{2}$

Key Concept

Adding Probabilities of Mutually Exclusive Events

Explanation

To find the probability of a compound event where no more than one of several mutually exclusive outcomes occurs, we add the probabilities of each individual outcome. In this case, we add the probabilities of flipping 0 heads and 1 head to find the probability of flipping no more than 1 head.

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