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In tossing four fair dice, what is the probability of tossing, at most 03 Jan 2021, 03:00
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Question Stats:

57% (02:46) correct 43% (02:22) wrong based on 150 sessions

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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
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D
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In tossing four fair dice, what is the probability of tossing, at most 03 Jan 2021, 04:16
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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
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In tossing four fair dice, what is the probability of tossing, at most 03 Jan 2021, 04:29
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\(P(Atmost-3)=P(No-3's)+4*P(one-3)=(\frac{5}{6})^4+4*\frac{1}{6}*(\frac{5}{6})^3=\frac{5^4+4*5}{6^4}=\frac{5^3(5+4)}{6^4}=\frac{5^3*9}{6^4}=\frac{5^3}{12*12}=\frac{125}{144}\)

Ans D
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In tossing four fair dice, what is the probability of tossing, at most 03 Jan 2021, 04:45
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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
[ Since each place can take 5 possible dice values - 1,2,4,5,6]

Combinations of tossing one 3 = 4*5*5*5=500
[ If i am tossing 3 on very first dice the other dice can't take 3. This means first dice can take 3 in only one way and other dice can take values other than 3 in 5 ways i.e. 1*5*5*5. Moreover, 3 can appear on second, third or fourth dice also. So total number of cases are multiplied by 4 i.e. 1*5*5*5*4=500 ]

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

Required probability= 1125/1296
= 125/144

Option D
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In tossing four fair dice, what is the probability of tossing, at most 17 Apr 2024, 12:18
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What should we the approach if the question asks for atmost two 3s
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In tossing four fair dice, what is the probability of tossing, at most 17 Apr 2024, 12:36
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VROOON153
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

What should we the approach if the question asks for atmost two 3s
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­
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 11 Jun 2025, 11:58
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