Question

A toddler is allowed to dress himself on Mondays, Wednesdays, and Fridays. For each of his shirt, pants, and shoes, he is equally likely to put it on correctly as incorrectly. Getting these articles of clothing on correctly are independent of each other. On the other days, the mother dresses the toddler with 100% accuracy. Given that the toddler is correctly dressed, what is the probability that today is Monday

Answer 1

Answer:

0.0286 = 2.86% probability that today is Monday.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Dressed correctly

Event B: Monday

Probability of being dressed correctly:

100% = 1 out of 4/7(mom dresses).

(0.5)^3 = 0.125 out of 3/7(toddler dresses himself). So

$P(A) = 0.125\frac{3}{7} + \frac{4}{7} = \frac{0.125*3 + 4}{7} = \frac{4.375}{7} = 0.625$

Probability of being dressed correctly and being Monday:

The toddler dresses himself on Monday, so (0.5)^3 = 0.125 probability of him being dressed correctly, 1/7 probability of being Monday, so:

$P(A \cap B) = 0.125\frac{1}{7} = 0.0179$

What is the probability that today is Monday?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0179}{0.625} = 0.0286$

0.0286 = 2.86% probability that today is Monday.