What is P (a n b) + P (a n b') + P (a' n b) =? And how?

 What is P (a n b) + P (a n b') + P (a' n b) =? And how?👇💝💝

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Full math Solution 👇👇👇👇

To determine $𝑃(𝐴\cap 𝐵)+𝑃(𝐴\cap {𝐵}^{\mathrm{\prime }})+𝑃({𝐴}^{\mathrm{\prime }}\cap 𝐵)$, let's use the properties of probability and set theory.

Firstly, let's understand the notation:

• $𝐴$ and $𝐵$ are events.
• ${𝐴}^{\mathrm{\prime }}$ is the complement of event $𝐴$ (i.e., ${𝐴}^{\mathrm{\prime }}$ happens when $𝐴$ does not happen).
• ${𝐵}^{\mathrm{\prime }}$ is the complement of event $𝐵$ (i.e., $𝐵$ happens when $𝐵$ does not happen).
• $\cap$ denotes the intersection of two events (i.e., both events occur).

Let's decompose the events within the probability space:

1. $𝐴\cap 𝐵$ is the event that both $𝐴$ and $𝐵$ occur.
2. $𝐴\cap {𝐵}^{\mathrm{\prime }}$ is the event that $𝐴$ occurs and $𝐵$ does not occur.
3. ${𝐴}^{\mathrm{\prime }}\cap 𝐵$ is the event that $𝐴$ does not occur and $𝐵$ occurs.

Now, let's visualize these events using a Venn diagram of events $𝐴$ and $𝐵$:

• The entire sample space $𝑆$ can be divided into four mutually exclusive regions:
1. $𝐴\cap 𝐵$ - Both $𝐴$ and $𝐵$ occur.
2. $𝐴\cap {𝐵}^{\mathrm{\prime }}$ - $𝐴$ occurs and $𝐵$ does not occur.
3. ${𝐴}^{\mathrm{\prime }}\cap 𝐵$ - $𝐴$ does not occur and $𝐵$ occurs.
4. ${𝐴}^{\mathrm{\prime }}\cap {𝐵}^{\mathrm{\prime }}$ - Neither $𝐴$ nor $𝐵$ occurs.

These four regions cover the entire sample space, meaning: $𝑃(𝐴\cap 𝐵)+𝑃(𝐴\cap {𝐵}^{\mathrm{\prime }})+𝑃({𝐴}^{\mathrm{\prime }}\cap 𝐵)+𝑃({𝐴}^{\mathrm{\prime }}\cap {𝐵}^{\mathrm{\prime }})=1$

Since we want to find $𝑃(𝐴\cap 𝐵)+𝑃(𝐴\cap {𝐵}^{\mathrm{\prime }})+𝑃({𝐴}^{\mathrm{\prime }}\cap 𝐵)$, let's rearrange the equation to isolate the desired terms: $𝑃(𝐴\cap 𝐵)+𝑃(𝐴\cap {𝐵}^{\mathrm{\prime }})+𝑃({𝐴}^{\mathrm{\prime }}\cap 𝐵)=1-𝑃({𝐴}^{\mathrm{\prime }}\cap {𝐵}^{\mathrm{\prime }})$

Now, we observe that $𝑃({𝐴}^{\mathrm{\prime }}\cap {𝐵}^{\mathrm{\prime }})$ is the probability that neither $𝐴$ nor $𝐵$ occurs.

Thus, the answer is: $𝑃(𝐴\cap 𝐵)+𝑃(𝐴\cap {𝐵}^{\mathrm{\prime }})+𝑃({𝐴}^{\mathrm{\prime }}\cap 𝐵)=1-𝑃({𝐴}^{\mathrm{\prime }}\cap {𝐵}^{\mathrm{\prime }})$

This is the final result, and it shows that the sum of the probabilities of the events $𝐴\cap 𝐵$, $𝐴\cap {𝐵}^{\mathrm{\prime }}$, and ${𝐴}^{\mathrm{\prime }}\cap 𝐵$ is equal to 1 minus the probability of ${𝐴}^{\mathrm{\prime }}\cap {𝐵}^{\mathrm{\prime }}$, the event where neither $𝐴$ nor $𝐵$ occurs.