Question

Who eats breakfast? The two-way table describes the 595 students who responded to a school survey about eating breakfast. Suppose we select a student at random. Consider events B: eats breakfast regularly, and $M$: is male.

(a) Find $P(B|M)$ Explain what this value means.

(b) Find$P(M|B)$ Explain what this value means.

Short answer

Part (a)$P\left(B\right|M)=0.5938$

Part (b)$P\left(M\right|B)=0.6333$

Step-by-step solution

Part (a) Step 1. Given Information

The adjusted gross income (in thousands of dollars) reported on individual federal income tax returns in a recent year is distributed as follows:

Part (a) Step 2. Concept Used

The number of favorable outcomes divided by the total number of possible outcomes equals probability. As a result, the following is a definition of condition probability: $P\left(A\right|B)=\frac{P(AandB)}{P\left(B\right)}$

Part (a) Step 3. Calculation

The person eats breakfast on a regular basis, according to the inquiry. Now we must determine the likelihood that the result for "male" is correct.

Therefore,

$P(BandM)=\frac{favourableoutcomes}{possibleoutcome}=\frac{190}{595}$

$P\left(M\right)=\frac{favourableoutcomes}{possibleoutcomes}=\frac{190+130}{595}=\frac{320}{595}$

As a result, the conditional probability is:

$P\left(B\right|M)=\frac{P(BandM)}{P\left(M\right)}=\frac{190}{320}\approx 0.5938$

As a result, the likelihood that the result for "male" is $P\left(B\right|M)=0.5938$

Part (b) Step 1. Calculation

The person in issue is a man, according to the inquiry. Now we must determine the likelihood that the result for "eats breakfast on a regular basis" is correct. Therefore,

$P(BandM)=\frac{favorableoutcomes}{possibleoutcomes}=\frac{190}{595}$

$P\left(M\right)=\frac{favourableoutcomes}{possibleoutcomes}=\frac{190+130}{595}=\frac{320}{595}$

As a result, the conditional probability is:

$P(BandM)=\frac{favourableoutcomes}{possibleoutcomes}=\frac{190}{595}$

$P\left(B\right)=\frac{favourableoutcomes}{possibleoutcomes}=\frac{190+110}{595}=\frac{300}{595}$

As a result, the conditional probability is: $P\left(M\right|B)=\frac{P(BandM)}{P\left(B\right)}$

$P\left(M\right|B)=\frac{190/595}{300/595}=\frac{190}{300}\approx 0.6333$

As a result, there's a good chance that the result for "eats breakfast regularly" will be positive$P\left(M\right|B)=0.6333$