In tossing four fair dice, what is the probability of tossing, at most

Question

In tossing four fair dice, what is the probability of tossing, at most one 3?

A. 19/144
B. 125/324
C. 625/1296
D. 125/144
E. 143/144

harvardforsure · Answer

At most one 3

Let us consider one 3

The probability will be: (1/6*5/6*5/6*5/6) arranged in 4 ways.

Thus total of 500/1296

Consider zero 3

Probability of : (5/6^4)=625/1296

Total 1125/1296 - reducing we get 125/144.
Hence D

imeanup · Answer

P(Atmost-3)=P(No-3's)+4*P(one-3)=(\frac{5}{6})^4+4*\frac{1}{6}*(\frac{5}{6})^3=\frac{5^4+4*5}{6^4}=\frac{5^3(5+4)}{6^4}=\frac{5^3*9}{6^4}=\frac{5^3}{12*12}=\frac{125}{144}

Ans D

Shadyshades · Answer

Four fair dice are tossed. 
 The probability of tossing at most one 3 = the probability of tossing no 3 + the probability of tossing one 3

Combinations of tossing no 3 = 5*5*5*5 = 625

Combinations of tossing one 3 = 4*5*5*5=500

Total combinations=625+500=1125
Total number of cases= 6*6*6*6= 1296

Required probability = 1125/1296
= 125/144

Option D

VROOON153 · Answer

What should we the approach if the question asks for atmost two 3s

Bunuel · Answer

P(zero, one or two 3s) =

= 1- P(thee or four 3s) =

= 1 - (P(three 3s) + P(four 3s)) =

= 1 - ((\frac{1}{6})^3*\frac{5}{6}*\frac{4!}{3!} + (\frac{1}{6})^4) =

= \frac{425}{432}

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In tossing four fair dice, what is the probability of tossing, at most

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In tossing four fair dice, what is the probability of tossing, at most

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