Some examples problems — lesson. Mathematics State Board, Class 10.

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Example 1:

$P(A) = \frac{1}{2}$ ,

$P(B) = \frac{1}{3}$ and

$P(A \cap B) = \frac{1}{6}$ , then find the value of

$P(A \cup B)$ .

Solution:

Addition Theorem of Probability:

$A$ and

$B$ are any two non mutually exclusive events then:

$P(A \cup B)$

$=$

$P(A)$

$+$

$P(B)$

$-$

$P(A \cap B)$

Substitute the known values in the above theorem.

$P(A \cup B)$

$=$

$\frac{1}{2}$

$+$

$\frac{1}{3}$

$-$

$\frac{1}{6}$

$=$

$\frac{3 + 2 - 1}{6}$

$=$

$\frac{5 - 1}{6}$

$=$

$\frac{4}{6}$

$=$

$\frac{2}{3}$

Therefore, the value of

$P(A \cup B)$ is

$\frac{2}{3}$ .

Example 2:

Find the probability of getting a king or a black card or a jack card from a well shuffled deck of

$52$ cards.

Solution:

Let

$S$ be the sample space.

Here

$S$

$=$

$52\text{ cards}$ .

$\Rightarrow$

$n(S) = 52$ .

Let

$A$ be the event of getting a king card.

So,

$n(A) = 4$ .

Thus,

$P(A) = \frac{n(A)}{n(S)} = \frac{4}{52}$ .

Let

$B$ be the event of getting a black card.

So,

$n(B) = 13$ .

Thus,

$P(B) = \frac{n(B)}{n(S)} = \frac{13}{52}$ .

Let

$C$ be the event of getting a jack card.

So,

$n(C) = 4$ .

Thus,

$P(C) = \frac{n(C)}{n(S)} = \frac{4}{52}$ .

Here,

$P(A \cap B) =$ Probability of getting a black king.

So,

$P(A \cap B) = \frac{2}{52}$ .

Here,

$P(B \cap C) =$ Probability of getting a black jack.

So,

$P(B \cap C) = \frac{2}{52}$ .

Here,

$P(A \cap C) =$ Probability of getting a king and a jack.

So,

$P(A \cap C) = \frac{0}{52}$ .

Thus,

$P(A \cap B \cap C)$

$=$

$\frac{0}{52}$ .

Addition Theorem of Probability:

$A$ ,

$B$ and

$C$ are any three non mutually exclusive events then:

$P(A \cup B \cup C)$

$=$

$P(A)$

$+$

$P(B)$

$+$

$P(C)$

$-$

$P(A \cap B)$

$-$

$P(B \cap C)$

$-$

$P(A \cap C)$

$+$

$P(A \cap B \cap C)$

Substitute the known values in the above theorem.

$P(A \cup B \cup C)$

$=$

$\frac{4}{52}$

$+$

$\frac{13}{52}$

$+$

$\frac{4}{52}$

$-$

$\frac{2}{52}$

$-$

$\frac{2}{52}$

$-$

$\frac{0}{52}$

$+$

$\frac{0}{52}$

$=$

$\frac{21 - 4 + 0}{52}$

$=$

$\frac{17}{52}$

Hence, the value of

$P(A \cup B \cup C)$ is

$\frac{17}{52}$ .

Therefore, the probability of getting a king or a black card or a jack card from a well shuffled deck of

$52$ cards is

$\frac{17}{52}$ .

Example 3:

$P(A) = \frac{1}{2}$ ,

$P(B) = \frac{1}{3}$ and

$P(A \cap B) = \frac{1}{6}$ , then find the value of (i)

$P(\text{only }A)$ (ii)

$P(\text{only }B)$ .

Solution:

By the theorem, we have:

(i)

$P(A \cap \overline B)$

$=$

$P(\text{only A})$

$=$

$P(A) - P(A \cap B)$

(ii)

$P(\overline A \cap B)$

$=$

$P(\text{only B})$

$=$

$P(B) - P(A \cap B)$

(i)

$P(\text{only A})$

$=$

$\frac{1}{2} - \frac{1}{6}$

$=$

$\frac{3 - 1}{6}$

$=$

$\frac{2}{6}$

$=$

$\frac{1}{3}$

Therefore, the value of

$P(\text{only A})$ is

$\frac{1}{3}$ .

(ii)

$P(\text{only B})$

$=$

$\frac{1}{3} - \frac{1}{6}$

$=$

$\frac{2 - 1}{6}$

$=$

$\frac{1}{6}$

Therefore, the value of

$P(\text{only B})$ is

$\frac{1}{6}$ .

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4. Some examples problems

Theory: