MHT CET · Maths · Probability

If three dices are thrown then the probability that the sum of the numbers on their uppermost faces to be atleast 5 is

A $\frac{1}{53}$
B $\frac{53}{54}$
C $\frac{1}{54}$
D $\frac{52}{53}$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{53}{54}$

Step-by-step Solution

Detailed explanation

If there dice are drawn, then

n (\leq) = 6^{3} = 216

Let E the sum of the number on their uppermost faces are less than 5 as follows

(1,1, 1), (1,1, 2), (1,2, 1), (2,1, 1)

∴ n (E) = 4

Now, probability that the sum of the numbers on their uppermost faces to be atleast 5

= 1 - P (E_{2}) = 1 - \frac{4}{216} = 1 - \frac{1}{54} = \frac{53}{54}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app