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16 Apr 2026, 11:25
Kudos
Bookmarks
Bunuel, pls help! I have a doubt- shouldn't the favourable outcomes be 1x5x5= 25 and then 25 multiplied by 3! because the 3 numbers could be arranged in 3! ways..I don't understand why we just did only 3 because 3 can come in any of the 3 throws..
16 Apr 2026, 11:37
Kudos
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If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?
A. 25/216
B. 50/216
C. 25/72
D. 25/36
E. 5/6
The probability is 3 * 1/6 * 5/6 * 5/6 = 25/72.
To get exactly one 3 in three rolls, one roll must be a 3 and the other two rolls must be not 3.
That is why we use:
1/6 for the roll that is a 3
5/6 for each roll that is not a 3.
Now count where the single 3 can appear:
3, not 3, not 3
not 3, 3, not 3
not 3, not 3, 3
So there are 3 possible positions for that one 3. Each of those cases has probability
1/6 * 5/6 * 5/6
Therefore total probability =
3 * 1/6 * 5/6 * 5/6 = 25/72.
Answer: C.
A. 25/216
B. 50/216
C. 25/72
D. 25/36
E. 5/6
The probability is 3 * 1/6 * 5/6 * 5/6 = 25/72.
To get exactly one 3 in three rolls, one roll must be a 3 and the other two rolls must be not 3.
That is why we use:
1/6 for the roll that is a 3
5/6 for each roll that is not a 3.
Now count where the single 3 can appear:
3, not 3, not 3
not 3, 3, not 3
not 3, not 3, 3
So there are 3 possible positions for that one 3. Each of those cases has probability
1/6 * 5/6 * 5/6
Therefore total probability =
3 * 1/6 * 5/6 * 5/6 = 25/72.
Answer: C.
16 Apr 2026, 11:38
Kudos
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Isha2701
You do not multiply by 3! because the two “not 3” rolls are already being counted separately in that 5 * 5, so their order is already built in. The only extra choice left is where the single 3 goes, and that gives 3 positions, not 6. So the correct count is 25 * 3 = 75, not 25 * 6.
