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Bunuel
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Two dice, each numbered 1 to 6, are tossed. What is the probability th 25 Jun 2022, 10:45
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C
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15% (low)

Question Stats:

81% (01:11) correct 19% (01:04) wrong based on 139 sessions

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Two dice, each numbered 1 to 6, are tossed. What is the probability that at least one show a "4" ?

A. 1/36
B. 4/36
C. 11/36
D. 25/36
E. 35/36
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C
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Two dice, each numbered 1 to 6, are tossed. What is the probability th 25 Jun 2022, 10:56
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Bunuel
Two dice, each numbered 1 to 6, are tossed. What is the probability that at least one show a "4" ?

A. 1/36
B. 4/36
C. 11/36
D. 25/36
E. 35/36
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We could find 44 + 4X + X4...or we could find 1-XX.

1-(5/6)(5/6) = 1-(25/36) = 11/36

Answer choice C.

A common mistake in going this route is coming up with the idea to use the 1-minus technique, doing all the math right, finding the probability of XX,...and then forgetting to subtract from 1... :facepalm_man: As soon as I see that there's going to be some little step at the end, I flip my computer mouse over. That way, when I go to click an answer choice, the upside down mouse serves as a reminder to make sure I did that last little step.
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Two dice, each numbered 1 to 6, are tossed. What is the probability th 25 Jun 2022, 14:50
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When the sets are mutually exclusive (no intersection) you can find what the author doesn’t want and subtract from the sample set. It’s easier and faster.
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Two dice, each numbered 1 to 6, are tossed. What is the probability th 16 Dec 2024, 19:38
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Two dice, each numbered 1 to 6, are tossed.

As we are rolling 2 dice => Number of cases = \(6^2\) = 36

What is the probability that at least one show a "4" ?

P(At least one show a "4") = 1 - P(None show a "4") = 1 - \(\frac{5}{6}\) * \(\frac{5}{6}\)
[ As we can get any number except 4 (1,2,3,5,6) out of the possible 6 cases from 1 to 6 in each roll ]

= 1 - \(\frac{25}{36}\) = \(\frac{11}{36}\)

So, Answer will be C
Hope it helps!

Playlist on Solved Problems on Probability here

Watch the following video to MASTER Dice Rolling Probability Problems

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Bunuel
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