The problem provides a frequency distribution table of marks obtained by students. Part (a) requires constructing a cumulative frequency table, drawing an ogive, and using the ogive to determine the 40th percentile. Part (b) asks for the probability of choosing a student who did not get distinction, given that 20% of the students had distinction. Problem 11 involves calculating probabilities related to drawing balls from a box containing black, red, and white balls without replacement. We are going to solve problem 11.

1. Problem Description

The problem provides a frequency distribution table of marks obtained by students. Part (a) requires constructing a cumulative frequency table, drawing an ogive, and using the ogive to determine the 40th percentile. Part (b) asks for the probability of choosing a student who did not get distinction, given that 20% of the students had distinction. Problem 11 involves calculating probabilities related to drawing balls from a box containing black, red, and white balls without replacement. We are going to solve problem

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2. Solution Steps

Problem

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(a) Probability the first ball is red.

There are 25 balls in total, and 7 are red. Therefore, the probability of drawing a red ball on the first draw is:

$P(\text{first ball is red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{7}{25}$

(b) Probability that both balls are white or black.

First, we need to determine the number of white balls.

Total balls = 25

Black balls = 10

Red balls = 7

White balls = 25 - 10 - 7 = 8

We need to find the probability that both balls are either white or black. This can happen in two mutually exclusive ways: both balls are white, or both balls are black, or one is white and one is black.

Case 1: Both balls are white.

$P(\text{both white}) = \frac{8}{25} \times \frac{7}{24} = \frac{56}{600}$

Case 2: Both balls are black.

$P(\text{both black}) = \frac{10}{25} \times \frac{9}{24} = \frac{90}{600}$

Case 3: One ball is white and one is black

$P(\text{first ball is white, second is black}) = \frac{8}{25} \times \frac{10}{24} = \frac{80}{600}$

$P(\text{first ball is black, second is white}) = \frac{10}{25} \times \frac{8}{24} = \frac{80}{600}$

$P(\text{one ball is white and one is black}) = \frac{80}{600} + \frac{80}{600} = \frac{160}{600}$

The question asks for the probability that both balls are white or black.

$P(\text{both balls are white or black}) = P(\text{both white}) + P(\text{both black})$

$= \frac{56}{600} + \frac{90}{600} = \frac{146}{600} = \frac{73}{300}$

(c) Probability that none of the balls is white.

This means both balls are either red or black.

Total number of non-white balls = 10 (black) + 7 (red) = 17

$P(\text{none is white}) = P(\text{both are red or black}) = \frac{17}{25} \times \frac{16}{24} = \frac{272}{600} = \frac{68}{150} = \frac{34}{75}$

3. Final Answer

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1. (a) $\frac{7}{25}$

(b) $\frac{73}{300}$

(c) $\frac{34}{75}$