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11 Aug 2017, 09:05
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7/12
As 1-[P(not getting 3) x P(not getting head)]===> 1- [5/6 x 1/2] ===> 7/12.
As 1-[P(not getting 3) x P(not getting head)]===> 1- [5/6 x 1/2] ===> 7/12.
13 Mar 2021, 08:45
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A fair die is rolled once and a fair coin is flipped once. What is the probability that either the die will land on 3 or that the coin will land on heads?
P(die got 3) = 1/6, P(coin got head)= 1/2
So, P(die got 3 or coin got head ) = 1/6 +1/2 - 1/6*1/2 = 7/12
It is C.
P(die got 3) = 1/6, P(coin got head)= 1/2
So, P(die got 3 or coin got head ) = 1/6 +1/2 - 1/6*1/2 = 7/12
It is C.
21 Jul 2025, 08:35
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A fair die is rolled once and a fair coin is flipped once. What is the probability that either the die will land on 3 or that the coin will land on heads?
P(H) = \(\frac{1}{2}\) (As we have 1 case out of 2 (head or tail) in which we can get a head)
P(3) = \(\frac{1}{6}\) (As we have 1 case (getting a 3) out of 6 (1, 2, 3, 4 ,5, 6) in which we can get a 3)
P getting either a head or getting 3 = P(Getting a Head) + P(Getting a 3) - P(Getting both H and 3) = P(H) + P(3) - P(H)*P(3) = \(\frac{1}{2}\) + \(\frac{1}{6}\) - \( \frac{1}{2} * \frac{1}{6}\) = \(\frac{6 + 2 - 1}{12}\) = \(\frac{7}{12}\)
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
P(H) = \(\frac{1}{2}\) (As we have 1 case out of 2 (head or tail) in which we can get a head)
P(3) = \(\frac{1}{6}\) (As we have 1 case (getting a 3) out of 6 (1, 2, 3, 4 ,5, 6) in which we can get a 3)
P getting either a head or getting 3 = P(Getting a Head) + P(Getting a 3) - P(Getting both H and 3) = P(H) + P(3) - P(H)*P(3) = \(\frac{1}{2}\) + \(\frac{1}{6}\) - \( \frac{1}{2} * \frac{1}{6}\) = \(\frac{6 + 2 - 1}{12}\) = \(\frac{7}{12}\)
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
