A dice is thrown once, what is the probability of getting a prime number? a. 1/3 b. 6/25 c. 1/2 d. 1/4

A dice is thrown once, what is the probability of getting a prime number?
a. 1/3
b. 6/25
c. 1/2
d. 1/4
by

1 Answer

Answer :

c. 1/2
by

Related questions

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Last Answer : d. 1/2

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Last Answer : Let A : Getting an odd number on first die; B : Getting a multiple of 3 on second die, thenA = {1, 3, 5}, B = {3, 6} ∴ P(A) = \(rac{3}{6}=rac{1}{2}\), P(B) = \(rac{2}{6}=rac{1}{3}\) ... B are independent∴ Required probability = P (A) . P (B) = \(rac{1}{2}\) x \(rac{1}{3}\) = \(rac{1}{6}\)

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Last Answer : Answer: A) Let the event of getting a greater number on the first die be G. There are 4 ways to get a sum of 9 when two dice are rolled = {(3,6),(4,5),(5,4), (6,3)}. And there are two ways where the number on the ... Now, P(G) = P(G sum equals 9)/P(sum equals 9) = (2/36)/(4/36) = 2/4 =>1/2

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Last Answer : d. 5/18

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Last Answer :  Answer: C)  Clearly, n(S) = (6 x 6) = 36. Let E = Event that the sum is a composite number Then E= { (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 5), (3, 3), (3, 6), (4, 2), (4,4),(4, 5), (4, 6), ( ... 5,3),(5,4),(5,5),(6,2),(6,3),(6,4),(6,6) } n(E) = 21 P(E) = n(E)/n(S) = 21/36 = 7/12.

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Last Answer : (c) \(rac{7}{12}\)Total number of ways in which 2 dice are rolled = 6 6 = 36 ⇒ n(S) = 36 Let A : Event of rolling an even number of 1st dice B : Event of rolling a total of 7 ⇒ A = {(2, 1), (2, 2) , (2, 6), (4 ... (rac{18}{36}\) + \(rac{6}{36}\) - \(rac{3}{36}\) = \(rac{21}{36}\) = \(rac{7}{12}\).

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