When a dice is rolled, what is the number of possible outcomes of obtaining an even number?

When a dice is rolled, what is the number of possible outcomes of obtaining an even number?
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When a dice is rolled, what is the number of possible outcomes of obtaining an even number?
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When a dice is rolled, what are all the possible outcomes?

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A dice is rolled number of times and its outcomes are recorded as below : -Maths 9th

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Last Answer : Total number of outcomes = 250 Total number of outcomes of getting odd numbers = 35 + 50 + 53 = 138 P (getting an odd number) = 138 / 250 = 69 / 125

A dice is rolled number of times and its outcomes are recorded as below : -Maths 9th

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Last Answer : Total number of outcomes = 250 Total number of outcomes of getting odd numbers = 35 + 50 + 53 = 138 P (getting an odd number) = 138 / 250 = 69 / 125

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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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Last Answer : (d) \(rac{1}{36}\)Total number of outcomes when three identical dice are rolled, n(S) = 6 6 6 = 216 Let A : Event of rolling same number or each dice ⇒ A = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), ... 6)} ⇒ n(A) = 6 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{6}{216}\) = \(rac{1}{36}\).

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Last Answer : When two unbiased dice are rolled, the possible out comes are∴ n(S) = 36 Let A : getting a multiple of 2 on one die and a multiple of 3 on the other die. ⇒ A = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), ( ... (3, 6), (6, 2), (6, 4)} ⇒ n(A) = 11∴ P(A) = \(rac{n(A)}{n(S)} =rac{11}{36}.\)

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Last Answer : (c) \(rac{11}{36}\)Total number of outcomes when two identical dice are rolled, n(S) = 6 6 = 36 Let A : Event of rolling a multiple of 2 on one die and a multiple of 3 on the other die ⇒ A = {(2, 3), (2, 6), (4, 3), (4, ... , 4), (3, 6)} ⇒ n(A) = 11 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{11}{36}\).

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Two dice are thrown. Find the probability of getting an odd number on the first die and a multiple of 3 on the other. -Maths 9th

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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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Consider an experiment of tossing two fair dice, one black and one red. What is the probability that the number on the black die divides the number on red die ? (A) 22 / 36 (B) 12 / 36 (C) 14 / 36 (D) 6 / 36

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Two dice are rolling simultaneously .What is the probability that the sum of the number on the two faces is divided by 5 Or 7. A) 13/36 B) 14/36 C) 11/36 D) 9/36

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Last Answer :  Answer: C) Clearly, n(S) = 6 x 6 = 36 Let E be the event that the sum of the numbers on the two faces is divided by 5or 7. Then,E = {(1,4),(1,6),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(4,6),(5,2),(5,5),(6,1),(6,4)} n(E) = 11. Hence, P(E) = n(E)/n(S) = 11/36

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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.

When two dice are thrown, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 9. A) 1/2 B) 1/5 C) 2/5 D) 4/2

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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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Two dice are thrown simultaneously 500 times. -Maths 9th

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Last Answer : (i) P (getting a sum more than 10) = P (getting a sum of 11) + P (getting a sum of 12) = 28/500 + 15/500 = 28 + 15/500 = 43/500 = 0.869 = 0.09 (ii) P (getting a sum less than or equal to 5) = P ( ... + P (getting a sum of 10) + P (getting a sum of 11) = 53/500 + 46/500 + 28/500 = 127/500 = 0.254

A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is -Maths 9th

Description : A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is -Maths 9th

Last Answer : (c) \(rac{84 imes5^6}{6^{10}}\)In the first nine throws we should have three sixes and six non-sixes and a six in the tenth throw and thereafter whatever face appears, it doesn't matter. ∴ Required probability = 9C3 \(\bigg(rac{1} ... x 1 x 1 ............x 1 {10 times} = \(rac{84 imes5^6}{6^{10}}\).