Question

Two fair dice are thrown simultaneously. The probability of getting the sum of numbers shown on the dice is either divisible by 2 or divisible by 5, is

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Answer 1

Answer:

Calculate the probabilities of each number possible with 2 dice, e.g. there is only 1 way of getting a 2, 2 ways of getting 3, 3 ways of getting a 4, 4 ways of 5 up to 6 ways of getting 7 (1,6, 2,5, 3,4, 4,3, 5,2, 6,1) and this reduces until there is only 1 way for a 12. The numbers that are divisible by 5 are 5 and 10. Hence the number of ways these can be thrown are 4 for the 5 and 3 for 10, a total of 7 chances. Add the total number of ways the dice could fall (1+2+3+4+5+6+5+4+3+2+1) =36. So the probability that you will throw a number that is divisible by 5 is 7/36 or 0.194444444.

Answer 2

SOLUTION

TO DETERMINE

Two fair dice are thrown simultaneously. The probability of getting the sum of numbers shown on the dice is either divisible by 2 or divisible by 5,

EVALUATION

Here Two fair dice are thrown simultaneously

So the event points are

= { (1,1), (1,2), (1,3), (1,4),(1,5),(1,6),(2,1), (2,2), (2,3), (2,4),(2,5),(2,6),(3,1), (3,2), (3,3), (3,4),(3,5),(3,6),(4,1), (4,2), (4,3), (4,4),(4,5),(4,6),(5,1), (5,2), (5,3), (5,4),(5,5),(5,6),(6,1), (6,2), (6,3), (6,4),(6,5),(6,6)}

So the total number of possible outcomes = 36

Let A & B be the events that the sum of numbers shown on the dice is divisible by 2 and divisible by 5 respectively

If the sum is divisible by 2 then the sum can 2 , 4, 6, 8, 10, 12

When sum is 2 points are (1,1)

When sum is 4 points are (1,3),(2,2),(3,1)

When sum is 6 points are (1,5),(2,4),(3,3),(4,2),(5,1)

When sum is 8 points are (2,6),(3,5),(4,4),(5,3),(6,2)

When sum is 10 points are (4,6),(5,5),(6,4)

When sum is 12 points are (6,6)

So the total number of possible outcomes for the event A is 18

If the sum is divisible by 5 then the sum can be 5 & 10

When sum is 5 points are (1,4),(2,3),(3,2),(4,1)

When sum is 10 points are (4 ,6),(5,5),(6,4)

So the total number of possible outcomes for the event B is 7

Now A ∩ B = The event that the sum is divisible by both 2 & 5

So the total number of possible outcomes for the event A ∩ B is 3

∴ P(A)

$\displaystyle \sf{ = \frac{18}{36} }$

∴ P(B)

$\displaystyle \sf{ = \frac{7}{36} }$

P(A∩B) = 3/36

Hence the probability that the sum of numbers shown on the dice is either divisible by 2 or divisible by 5

= P(A∪B)

= P(A) + P(B) - P(A∩B)

$\displaystyle \sf{ = \frac{18}{36} + \frac{7}{36} - \frac{3}{36} }$

$\displaystyle \sf{ = \frac{18 + 7 - 3}{36} }$

$\displaystyle \sf{ = \frac{22}{36} }$

$\displaystyle \sf{ = \frac{11}{18} }$

FINAL ANSWER

$\displaystyle \sf{The \: required \: probability = \frac{11}{18} }$

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