Math  /  Data & Statistics

QuestionThe table gives a set of outcomes and their probabilities. Let AA be the event "the outcome is divisible by 33^{\prime \prime}. Find P(A)P(A). \begin{tabular}{|c|c|} \hline Outcome & Probability \\ \hline 1 & 0.11 \\ \hline 2 & 0.14 \\ \hline 3 & 0.03 \\ \hline 4 & 0.16 \\ \hline 5 & 0.05 \\ \hline 6 & 0.09 \\ \hline 7 & 0.19 \\ \hline 8 & 0.23 \\ \hline \end{tabular} \square

Studdy Solution

STEP 1

What is this asking? What's the chance that a randomly chosen outcome from this table is divisible by 33? Watch out! Don't forget to only consider the outcomes divisible by 33, and make sure to add the probabilities correctly!

STEP 2

1. Identify outcomes divisible by 3
2. Sum their probabilities

STEP 3

Let's **scan** our table and **spot** the outcomes divisible by 33.
We've got 33 and 66!
These are the only numbers in our table that you can divide by 33 and get a whole number.

STEP 4

The probability of getting a 33 is 0.030.03.
The probability of getting a 66 is 0.090.09.
To get the probability of *either* event happening, we **add** their probabilities!

STEP 5

So, P(A)P(A), the probability of the outcome being divisible by 33, is the probability of getting a 33 *plus* the probability of getting a 66.
Let's **calculate** that: P(A)=0.03+0.09=0.12 P(A) = 0.03 + 0.09 = 0.12

STEP 6

The probability of the outcome being divisible by 33 is 0.120.12.

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