Question

Three cooks, A, B and C bake a special kind of cake, and with respective probabilities 0.02, 0.03, and 0.05 it fails to rise. In the restaurant where they work, A bake 50 percent of these cakes, B 30 percent and C 20 percent. What proportion of failures is caused by A

Answer 1

Answer on Question # 45525 – Math – Statistics and Probability

Three cooks, A, B and C bake a special kind of cake, and with respective probabilities 0.02, 0.03, and 0.05 it fails to rise. In the restaurant where they work, A bake 50 percent of these cakes, B 30 percent and C 20 percent. What proportion of failures is caused by A.

Solution:

We start to solve with definition of the probability events applying to our problem.

The probability of an event A occurring when it is known that some event A has occurred is called a conditional probability and is denoted by $P(A|B)$. The symbol $P(A|B)$ is usually read the probability that A occurs given that B occurs or simply the probability of A given B.

The conditional probability of A, given B, denoted by $P(A|B)$, is defined the following formula.

$$P(A|B) = \frac{P(A \cap B)}{P(B)} \text{ if } P(B) > 0$$

Note the given values accordingly to the condition of the task. The cook A bake a $50\%$ of these cakes with probability $P(A) = 0.5$, the cook B bake a $30\%$ of cakes with probability $P(B) = 0.3$ and the cook C bake a $20\%$ of cakes with probability $P(C) = 0.2$.

Let F be the event that the cake fails to rise. Then we can write the probability with takes into account this condition.

$$P(F|A) = 0.02, \quad P(F|B) = 0.03 \text{ and } P(F|C) = 0.05$$

To solve our problem we apply the Bayes' Theorem. Accordingly to the theorem it should be noted.

Let the events $A_1, A_2, \ldots, A_k$ form a partition of the space $S$ such that $P(A_j) > 0$, for $j = 1, \ldots, k$, and let B be any event such that $P(B) > 0$. Then, for $j = 1, \ldots, k$,

$$P(A_j|B) = \frac{P(A_j)P(B|A_j)}{P(B)} = \frac{P(A_j)P(B|A_j)}{\sum_{i=1}^{k} P(A_i)P(B|A_i)}$$

Apply the formula noted above to solve our problem.

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

Where $P(F)$ is equal to the following formula.

$$P(F) = P(A \cap F) + P(B \cap F) + P(C \cap F) = P(A)P(F|A) + P(B)P(F|B) + P(C)P(F|C)$$

According to the condition of the task we have all data, so we can substitute into the formula noted above.

$$\mathrm{P}(\mathrm{F}) = (0.5 \cdot 0.02) + (0.3 \cdot 0.03) + (0.2 \cdot 0.05) = 0.01 + 0.009 + 0.01 = 0.029$$

Now we can substitute the obtained value into the formula. We know that

$$\mathrm{P}(\mathrm{A} \cap \mathrm{F}) = \mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{F} | \mathrm{A}) = 0.01$$

So, we can find the value of $\mathrm{P}(\mathrm{A} | \mathrm{F})$.

$$\mathrm{P}(\mathrm{A} | \mathrm{F}) = \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})} = \frac{0.01}{0.029} \approx 0.344827$$

Finally we can write that $\mathrm{P}(\mathrm{A} | \mathrm{F}) = 0.34482$

Answer: The proportion of failures is caused by A is equal to $\mathrm{P}(\mathrm{A} | \mathrm{F}) = 0.34482$ (approximately 34%).

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