CUET Mathematics 2024 - Two dice are thrown simultaneously. If X denotes the number of fours, then the expectation of X will be : | PYQs + Solutions

CUET Mathematics 2024

Algebra

Probability

Medium

Two dice are thrown simultaneously. If X denotes the number of fours, then the expectation of X will be :

\frac{5}{9}

\frac{1}{3}

\frac{4}{7}

\frac{3}{8}

✅ Correct Option: 2

Faster method using linearity of expectation:

For each die, the expected number of fours is

\dfrac{1}{6}

Since we have two independent dice:

E(X) = E(X_1) + E(X_2) = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}

Therefore, the expectation of

X

\dfrac{1}{3}

Alternate Method:

Let

X

denote the number of fours that appear when two dice are thrown.

We need to find the expected value

E(X)

Possible values of

X

are 0, 1, and 2.

When rolling a single die:

\newline

- Probability of rolling a 4 =

\dfrac{1}{6}

\newline

- Probability of not rolling a 4 =

\dfrac{5}{6}

Calculating the probability for each value of

X

For

X = 0

(no fours appear):

P(X = 0) = \dfrac{5}{6} \times \dfrac{5}{6} = \dfrac{25}{36}

For

X = 1

(exactly one four appears):

P(X = 1) = \left(\dfrac{1}{6} \times \dfrac{5}{6}\right) + \left(\dfrac{5}{6} \times \dfrac{1}{6}\right) = \dfrac{10}{36}

For

X = 2

(both dice show four):

P(X = 2) = \dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}

Calculating the expectation using

E(X) = \sum x \times P(X = x)

E(X) = 0 \times P(X = 0) + 1 \times P(X = 1) + 2 \times P(X = 2)

E(X) = 0 \times \dfrac{25}{36} + 1 \times \dfrac{10}{36} + 2 \times \dfrac{1}{36}

E(X) = \dfrac{10}{36} + \dfrac{2}{36} = \dfrac{12}{36} = \dfrac{1}{3}

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