CBSE Sample Paper 2020-21 · Part B: Section III · 2 marks

Q28: Distribution of the number of milk chocolates drawn, or $P(\bar E \mid \bar F)$

Sample 2020-21Ch 13 · Probability

The question

(a) A refrigerator box contains 2 milk chocolates and 4 dark chocolates. Two chocolates are drawn at random. Find the probability distribution of the number of milk chocolates. What is the most likely outcome?

(b) Given that E and F are events such that $P(E) = 0.8$ ⁠, $P(F) = 0.7$ ⁠, $P(E \cap F) = 0.6$ ⁠. Find $P(\bar E \mid \bar F)$

Part (a) uses a random variable and its probability distribution. This topic is no longer in the CBSE syllabus (2025-26).

Exam answer

(a)

Let $X$ denote the number of milk chocolates drawn. $X$ can be 0, 1 or 2.

\begin{aligned} P(X = 0) &= \frac46 \times \frac35 = \frac{12}{30} \\ P(X = 1) &= \left(\frac26 \times \frac45\right) \times 2 \\ &= \frac{16}{30} \\ P(X = 2) &= \frac26 \times \frac15 = \frac{2}{30} \end{aligned}

Why?

The two chocolates are drawn one after the other without replacement, so after the first draw only 5 are left.

$X = 0$ ⁠: dark, then dark: 4 of the 6 are dark, then 3 of the 5 left.

$X = 1$ ⁠: milk then dark, or dark then milk. Each order has probability $\frac26 \times \frac45$ ( $\frac46 \times \frac25$ is the same number), so we double it.

$X = 2$ ⁠: milk, then milk: 2 of 6, then 1 of 5.

Multiply tops together and bottoms together:

\begin{aligned} \frac46 \times \frac35 &= \frac{4 \times 3}{6 \times 5} = \frac{12}{30} \\[4pt] \frac26 \times \frac45 &= \frac{2 \times 4}{6 \times 5} = \frac{8}{30} \\ \frac{8}{30} \times 2 &= \frac{16}{30} \\[4pt] \frac26 \times \frac15 &= \frac{2 \times 1}{6 \times 5} = \frac{2}{30} \end{aligned}

\begin{array}{c|c|c|c} X & 0 & 1 & 2 \\ \hline P(X) & \frac{12}{30} & \frac{16}{30} & \frac{2}{30} \end{array}

Why?

A probability distribution lists every value of $X$ with its probability. Check: the probabilities add up to 1, since $12 + 16 + 2 = 30$ ⁠.

\begin{aligned} &\frac{12}{30} + \frac{16}{30} + \frac{2}{30} \\ &= \frac{12 + 16 + 2}{30} \\ &= \frac{28 + 2}{30} = \frac{30}{30} = 1 \end{aligned}

The most likely outcome is $X = 1$ ⁠: getting one chocolate of each type.

Why?

The most likely outcome is the value with the largest probability, here $\frac{16}{30}$ ⁠.

(b)

\begin{aligned} &P(\bar E \mid \bar F) \\ &= \frac{P(\bar E \cap \bar F)}{P(\bar F)} \\ &= \frac{P(\overline{E \cup F})}{P(\bar F)} \\ &= \frac{1 - P(E \cup F)}{1 - P(F)} \quad\ldots(\text{i}) \end{aligned}

Why?

Conditional probability:

P(A \mid B) = \frac{P(A \cap B)}{P(B)}

By De Morgan's law, "not E and not F" is "not (E or F)": $\bar E \cap \bar F = \overline{E \cup F}$ ⁠. The probability of a complement is 1 minus the probability of the event.

\begin{aligned} &P(E \cup F) \\ &= P(E) + P(F) - P(E \cap F) \\ &= 0.8 + 0.7 - 0.6 = 0.9 \end{aligned}

Why?

This is the addition rule. $P(E \cap F)$ is subtracted because the outcomes in both E and F are counted twice in $P(E) + P(F)$ ⁠.

\begin{aligned} 0.8 + 0.7 &= 1.5 \\ 1.5 - 0.6 &= 0.9 \end{aligned}

Substituting in (i),

P(\bar E \mid \bar F) = \frac{1 - 0.9}{1 - 0.7} = \frac{0.1}{0.3}

Why?

Put $P(E \cup F) = 0.9$ and $P(F) = 0.7$ into (i):

\begin{aligned} 1 - 0.9 &= 0.1 \\ 1 - 0.7 &= 0.3 \end{aligned}

Multiply the top and the bottom by 10: $\frac{0.1}{0.3} = \frac{1}{3}$ ⁠.

\frac{0.1}{0.3} = \frac{0.1 \times 10}{0.3 \times 10} = \frac{1}{3}

$\therefore\ P(\bar E \mid \bar F) = \dfrac13$

Where this comes from

Chapter 13, Probability: Probability distribution; conditional probability.

Question type: (b) Given numbers

How to spot it: click to see

(b) Type 1. Given numbers: divide by the event after the bar

Spot it: the question gives $P(E)$ ⁠, $P(F)$ and $P(E\cap F)$ ⁠, and asks for $P(E|F)$ or $P(F|E)$ ⁠.

Here: The numbers are given, and $\bar F$ is after the bar, so divide by $P(\bar F) = 1 - 0.7 = 0.3$ ⁠. The top, $P(\bar E \cap \bar F)$ ⁠, is $1 - P(E \cup F) = 0.1$ from the addition rule.

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Q28: Distribution of the number of milk chocolates drawn, or P(Eˉ∣Fˉ)P(\bar E \mid \bar F)P(Eˉ∣Fˉ)

Q28: Distribution of the number of milk chocolates drawn, or $P(\bar E \mid \bar F)$