A probability problem by Aayush Gupta

Probability Level 3

What is the probability that after throwing 6 fair dice, all of the numbers facing up are different?

Express the answer as a decimal to the nearest thousandth.

The answer is 0.015.

1 solution

Aayush Gupta
Mar 29, 2014

We analyze the throws in order.

After the first throw, the second throw must be anything except for the first throw with a probability of 5/6.

The third throw must be anything but the second or first, with a probability of 4/6.

Continuing for every throw, we find the answer to be $\frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{1}{6} = \frac{5}{324}$

Rounding, we find the answer to be approximately $0.015$

Another way of thinking about it is the number of ways to arrange the faces on the die, and then divide by the number of ways to roll the die (i.e. $\dfrac{6!}{6^6}$ ).

Finn Hulse - 7 years, 1 month ago

what the hell...... I just wrote 0.0154 and forgot to round of till nearest 1000th and I thought that my thinking was wrong.....

Vighnesh Raut - 7 years ago

Same Here!!!! I used (6!/6^6) and forgot to round off !!!!

Kunal Gupta - 6 years, 9 months ago

too bad ...................... i am sorry to hear that

math man - 6 years, 9 months ago

I tried to solve it this way.

There are $6^{6}$ possible outcomes when throwing 6 dice. There are 6! ways of throwing 6 dice with different faces up. Hence, the probability is given by $\frac{6!}{6^{6}}$ , which is approximately 0.015.

Solomon Olayta - 5 years, 2 months ago

A probability problem by Aayush Gupta

The answer is 0.015.

1 solution

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