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Probability - Theorem of total probability

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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A set of events E1,E2,…,EnE_1, E_2, \dots, E_n is called a partition of the sample space SS if they are pairwise disjoint (Ei∩Ej=βˆ…E_i \cap E_j = \emptyset for iβ‰ ji \neq j), exhaustive (βˆͺi=1nEi=S\cup_{i=1}^n E_i = S), and each has a non-zero probability (P(Ei)>0P(E_i) > 0).

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The Theorem of Total Probability states that if E1,E2,…,EnE_1, E_2, \dots, E_n is a partition of SS, then for any event AA associated with SS, the probability P(A)P(A) is the weighted average of conditional probabilities P(A∣Ei)P(A|E_i).

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This theorem is often the first step in calculating posterior probabilities using Bayes' Theorem.

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To apply the theorem, identify the mutually exclusive events EiE_i (the 'causes' or 'paths') and the common event AA that can occur under each EiE_i.

πŸ“Formulae

P(A)=P(E1)P(A∣E1)+P(E2)P(A∣E2)+β‹―+P(En)P(A∣En)P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + \dots + P(E_n)P(A|E_n) outdoor

$$P(A) = \sum_{j=1}^{n} P(E_j)P(A|E_j)

$$P(A \cap E_i) = P(E_i)P(A|E_i)

πŸ’‘Examples

Problem 1:

A bag contains 44 red and 33 black balls. A second bag contains 22 red and 44 black balls. One bag is selected at random and a ball is drawn. Find the probability that the ball drawn is red.

Solution:

Let E1E_1 and E2E_2 be the events of selecting Bag I and Bag II respectively. Let AA be the event of drawing a red ball.

Since the bags are chosen at random: P(E1)=12P(E_1) = \frac{1}{2}, P(E2)=12P(E_2) = \frac{1}{2}

Probability of drawing a red ball from Bag I: P(A∣E1)=47P(A|E_1) = \frac{4}{7}

Probability of drawing a red ball from Bag II: P(A∣E2)=26=13P(A|E_2) = \frac{2}{6} = \frac{1}{3}

Using the Theorem of Total Probability: P(A)=P(E1)P(A∣E1)+P(E2)P(A∣E2)P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) P(A)=(12Γ—47)+(12Γ—13)P(A) = \left(\frac{1}{2} \times \frac{4}{7}\right) + \left(\frac{1}{2} \times \frac{1}{3}\right) P(A)=27+16P(A) = \frac{2}{7} + \frac{1}{6} P(A)=12+742=1942P(A) = \frac{12 + 7}{42} = \frac{19}{42}

Calculation of the numerator: 12+719\begin{array}{r} 12 \\ + 7 \\ \hline 19 \end{array}

Explanation:

We first define the partition of the sample space as choosing Bag I or Bag II. Then we calculate the conditional probability of drawing a red ball from each specific bag. Finally, we sum the products of the bag selection probability and the respective conditional probability.

Problem 2:

In a factory, machine M1M_1 produces 60%60\% of the items and machine M2M_2 produces 40%40\%. 2%2\% of the items produced by M1M_1 are defective, while 5%5\% of items from M2M_2 are defective. An item is chosen at random. Find the probability it is defective.

Solution:

Let E1E_1 be the event that the item is produced by M1M_1 and E2E_2 be the event it is produced by M2M_2. Let DD be the event that the item is defective.

P(E1)=60100=0.6P(E_1) = \frac{60}{100} = 0.6 P(E2)=40100=0.4P(E_2) = \frac{40}{100} = 0.4

P(D∣E1)=2100=0.02P(D|E_1) = \frac{2}{100} = 0.02 P(D∣E2)=5100=0.05P(D|E_2) = \frac{5}{100} = 0.05

By Total Probability Theorem: P(D)=P(E1)P(D∣E1)+P(E2)P(D∣E2)P(D) = P(E_1)P(D|E_1) + P(E_2)P(D|E_2) P(D)=(0.6)(0.02)+(0.4)(0.05)P(D) = (0.6)(0.02) + (0.4)(0.05) P(D)=0.012+0.020=0.032P(D) = 0.012 + 0.020 = 0.032

Sum calculation: 0.012+0.0200.032\begin{array}{r} 0.012 \\ + 0.020 \\ \hline 0.032 \end{array}

Explanation:

The event of being defective can happen via two paths: either from machine 1 or machine 2. The total probability is the sum of probabilities of being 'Defective and from M1M_1' and 'Defective and from M2M_2'.