If three dice are thrown together, then the probability of getting same digit on all the dice

Question

If three dice are thrown together, then the probability of getting same digit on all the dice (A)  (1/6)  (B)  (3/36)  (C)  (1/36)  (D)  (2/3)

Tardigrade Admin · Accepted Answer

Number of ways of getting the same digit on all the three dice
{(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5) and (6, 6, 6)}
Total number of ways

= 6^{3} = 216

Thus, required probability

= \frac{6}{216} = \frac{1}{36}

Q. If three dice are thrown together, then the probability of getting same digit on all the dice

$\frac{1}{6}$

$\frac{3}{36}$

$\frac{1}{36}$

$\frac{2}{3}$

Number of ways of getting the same digit on all the three dice
{(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5) and (6, 6, 6)}
Total number of ways $= 6^{3} = 216$
Thus, required probability $= \frac{6}{216} = \frac{1}{36}$

Q. If three dice are thrown together, then the probability of getting same digit on all the dice

61​

363​

361​

32​

Number of ways of getting the same digit on all the three dice {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5) and (6, 6, 6)} Total number of ways =63=216 Thus, required probability =2166​=361​

$\frac{1}{6}$

$\frac{3}{36}$

$\frac{1}{36}$

$\frac{2}{3}$

Number of ways of getting the same digit on all the three dice
{(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5) and (6, 6, 6)}
Total number of ways $= 6^{3} = 216$
Thus, required probability $= \frac{6}{216} = \frac{1}{36}$