A. Find the probability that the ball is White or red ii. Not white B. If A ∩ B = Ø on random experiment such that P(A)= 1/5 and P(B)= 1/3. Find the probability of the following events i. A or B happens ii. A and B happens iii. A or B not happens

I will solve the probability questions (16.A and 16.B) as they are complete. Questions B, C, and D regarding the pyramid cannot be solved without the initial description or diagram of the pyramid, which is missing from the image.

Question 16.A: A box containing 7 red balls, 4 blue balls and 10 white one. A ball is taken at random.

First, calculate the total number of balls: Total number of balls = $7 (red) + 4 (blue) + 10 (white) = 21$ balls.

i. Find the probability that the ball is White or red

Step 1: Identify the number of white and red balls. Number of white balls = 10 Number of red balls = 7

Step 2: Calculate the number of balls that are white or red. Number of (white or red) balls = $10 + 7 = 17$

Step 3: Calculate the probability. $P(White or Red) = \frac{Number of (white or red) balls}{Total number of balls} = \frac{17}{21}$ The probability that the ball is white or red is $\boxed{\frac{17}{21}}$.

ii. Find the probability that the ball is Not white

Step 1: Identify the number of non-white balls. Non-white balls are blue or red. Number of blue balls = 4 Number of red balls = 7 Number of (not white) balls = $4 + 7 = 11$

Step 2: Calculate the probability. $P(Not White) = \frac{Number of (not white) balls}{Total number of balls} = \frac{11}{21}$ The probability that the ball is not white is $\boxed{\frac{11}{21}}$.

Question 16.B: If $A \cap B = \emptyset$ on random experiment such that $P(A) = \frac{1}{5}$ and $P(B) = \frac{1}{3}$. Find the probability of the following events.

The condition $A \cap B = \emptyset$ means that events A and B are mutually exclusive.

i. A or B happens

Step 1: Use the formula for the probability of the union of two mutually exclusive events. For mutually exclusive events, $P(A or B) = P(A \cup B) = P(A) + P(B)$.

Step 2: Substitute the given probabilities and calculate. $P(A \cup B) = \frac{1}{5} + \frac{1}{3}$ To add the fractions, find a common denominator, which is 15. $P(A \cup B) = \frac{1 \times 3}{5 \times 3} + \frac{1 \times 5}{3 \times 5} = \frac{3}{15} + \frac{5}{15} = \frac{3+5}{15} = \frac{8}{15}$ The probability that A or B happens is $\boxed{\frac{8}{15}}$.

ii. A and B happens

Step 1: Understand what "A and B happens" means in terms of probability. This refers to the probability of the intersection of A and B, $P(A \cap B)$.

Step 2: Use the given information that $A \cap B = \emptyset$. Since the intersection of A and B is an empty set, it means the events cannot happen simultaneously. $P(A \cap B) = 0$ The probability that A and B happens is $\boxed{0}$.

iii. A or B not happens

Step 1: Understand what "A or B not happens" means. This refers to the complement of the event "A or B happens", which is $P((A \cup B)')$.

Step 2: Use the complement rule: $P(E') = 1 - P(E)$. $P((A \cup B)') = 1 - P(A \cup B)$

Step 3: Substitute the value of $P(A \cup B)$ from part 16.B.i. $P((A \cup B)') = 1 - \frac{8}{15}$ $P((A \cup B)') = \frac{15}{15} - \frac{8}{15} = \frac{15-8}{15} = \frac{7}{15}$ The probability that A or B not happens is $\boxed{\frac{7}{15}}$.