CBSE Class 12 Maths Exemplar Chapter 13: Probability NCERT Solutions

For the given loaded die, the probabilities of the outcomes are:

P(1) = P(2) = 0.2

P(3) = P(5) = P(6) = 0.1

P(4) = 0.3

The die is thrown two times. Let A be the event that the same number appears each time, and B be the event that the total score is 10 or more.

The event A consists of the outcomes where both throws result in the same number:

A = \{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\}

The probability of event A, P(A), is calculated as the sum of the probabilities of these individual outcomes. Since the throws are independent, the probability of an outcome (i, i) is P(i) * P(i).

P(A) = P(1) \cdot P(1) + P(2) \cdot P(2) + P(3) \cdot P(3) + P(4) \cdot P(4) + P(5) \cdot P(5) + P(6) \cdot P(6)

P(A) = (0.2 \times 0.2) + (0.2 \times 0.2) + (0.1 \times 0.1) + (0.3 \times 0.3) + (0.1 \times 0.1) + (0.1 \times 0.1)

P(A) = 0.04 + 0.04 + 0.01 + 0.09 + 0.01 + 0.01 = 0.20

The event B consists of the outcomes where the sum of the two throws is 10 or more. These outcomes are:

B = \{(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)\}

The probability of event B, P(B), is calculated as the sum of the probabilities of these outcomes:

P(B) = P(4) \cdot P(6) + P(6) \cdot P(4) + P(5) \cdot P(5) + P(5) \cdot P(6) + P(6) \cdot P(5) + P(6) \cdot P(6)

P(B) = (0.3 \times 0.1) + (0.1 \times 0.3) + (0.1 \times 0.1) + (0.1 \times 0.1) + (0.1 \times 0.1) + (0.1 \times 0.1)

P(B) = 0.03 + 0.03 + 0.01 + 0.01 + 0.01 + 0.01 = 0.10

Now, we find the intersection of events A and B, which means the outcomes where the same number is thrown each time AND the total score is 10 or more. These outcomes are:

A \cap B = \{(5, 5), (6, 6)\}

The probability of the intersection, P(A ∩ B), is:

P(A \cap B) = P(5) \cdot P(5) + P(6) \cdot P(6)

P(A \cap B) = (0.1 \times 0.1) + (0.1 \times 0.1) = 0.01 + 0.01 = 0.02

Two events A and B are independent if P(A \cap B) = P(A) \cdot P(B). Let's check this condition:

P(A) \cdot P(B) = 0.20 \times 0.10 = 0.02

Since P(A \cap B) = 0.02 and P(A) \cdot P(B) = 0.02, we have P(A \cap B) = P(A) \cdot P(B).

Therefore, events A and B are independent.

For a fair die, the probability of each outcome (1, 2, 3, 4, 5, 6) is \frac{1}{6}. The die is thrown two times. Let A be the event 'same number each time' and B be the event 'a total score is 10 or more'. The total number of possible outcomes when a die is thrown twice is n(S) = 6 \times 6 = 36.

The event A (same number each time) consists of the outcomes:

A = \{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\}

The number of favorable outcomes for A is n(A) = 6. The probability of A is:

P(A) = \frac{n(A)}{n(S)} = \frac{6}{36} = \frac{1}{6}

The event B (total score is 10 or more) consists of the outcomes:

B = \{(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)\}

The number of favorable outcomes for B is n(B) = 6. The probability of B is:

P(B) = \frac{n(B)}{n(S)} = \frac{6}{36} = \frac{1}{6}

The intersection of A and B (A \cap B) is the event where the same number is thrown each time AND the total score is 10 or more. The outcomes for A \cap B are:

A \cap B = \{(5, 5), (6, 6)\}

The number of favorable outcomes for A \cap B is n(A \cap B) = 2. The probability of A \cap B is:

P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{36} = \frac{1}{18}

To check for independence, we compare P(A \cap B) with the product P(A) \cdot P(B):

P(A) \cdot P(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}

Since P(A \cap B) = \frac{1}{18} and P(A) \cdot P(B) = \frac{1}{36}, we see that P(A \cap B)

eq P(A) \cdot P(B).

Therefore, for a fair die, events A and B are not independent.

We are given the probability that at least one of the two events A and B occurs, which is represented by P(A \cup B). We are also given the probability that both A and B occur simultaneously, which is represented by P(A \cap B).

Given:

P(A \cup B) = 0.6

P(A \cap B) = 0.3

We use the fundamental formula for the union of two events:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Substitute the given values into the formula:

0.6 = P(A) + P(B) - 0.3

To find the sum P(A) + P(B), we rearrange the equation:

P(A) + P(B) = 0.6 + 0.3

P(A) + P(B) = 0.9

Thus, the sum of the probabilities of events A and B is 0.9.

(Note: The additional lines in the source regarding P(\overline{A}) + P(\overline{B}) are not directly required to answer the question P(A) + P(B) but show a related calculation.)

Let R denote the event of drawing a red marble and B denote the event of drawing a black marble. The bag initially contains 5 red marbles and 3 black marbles, making a total of 8 marbles.

We are drawing three marbles one by one without replacement. We are given that the first marble drawn is red. We need to find the probability that at least one of the three marbles drawn is black, given that the first one is red.

Let E be the event that at least one of the three marbles drawn is black, given the first is red. This means that among the second and third draws, at least one must be black.

Since the first marble drawn is red, there are now 7 marbles left in the bag: 4 red and 3 black.

We can find the probability of the complementary event: that none of the remaining two marbles drawn (second and third) are black, meaning both are red. Let this complementary event be E'.

The probability of the second marble being red (given the first was red) is $\backslash frac\{4\}\{7\}$.

After drawing a second red marble, there are 6 marbles left: 3 red and 3 black.

The probability of the third marble being red (given the first two were red) is $\backslash frac\{3\}\{6\}\; =\; \backslash frac\{1\}\{2\}$.

So, the probability of drawing two red marbles in the second and third draws (event E') is:

$$P(E\text{'})\; =\; P(\backslash text\{2nd\; is\; Red\; |\; 1st\; is\; Red\})\; \backslash times\; P(\backslash text\{3rd\; is\; Red\; |\; 1st\; and\; 2nd\; are\; Red\})$$

$$P(E\text{'})\; =\; \backslash frac\{4\}\{7\}\; \backslash times\; \backslash frac\{3\}\{6\}\; =\; \backslash frac\{4\}\{7\}\; \backslash times\; \backslash frac\{1\}\{2\}\; =\; \backslash frac\{4\}\{14\}\; =\; \backslash frac\{2\}\{7\}$$

The probability of event E (at least one black marble in the second and third draws) is $1\; -\; P(E\text{'})$.

$$P(E)\; =\; 1\; -\; P(E\text{'})\; =\; 1\; -\; \backslash frac\{2\}\{7\}\; =\; \backslash frac\{5\}\{7\}$$

Therefore, the probability that at least one of the three marbles drawn is black, given the first marble is red, is $\backslash frac\{5\}\{7\}$.