Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let X is the random variable score obtained when a die is thrown twice.X = 1, 2, 3, 4, 5, 6
Here, S = {(1, 1), (1, 2), ( 2, 1), ( 2, 2), (1, 3), ( 2, 3), ( 3, 1), ( 3, 2), ( 3, 3), ..., ( 6, 6)}
$\begin{array}{l} ∴ P (X = 1) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \\ P (X = 2) = \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} = \frac{3}{36} \\ P (X = 3) = \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} = \frac{5}{36} \end{array}$
Similarly, $P (X = 4) = \frac{7}{36}, P (X = 5) = \frac{9}{36} and P (X = 6) = \frac{11}{36}$
So, the required distribution is,