Question

a dice is rolled twice. find the probability that 5 will not come up either time and 5 will come up exactly one time

Answer 1

Total event =6×6 =36
Events of getting 5 are = (5,1),( 5,2),( 5,3),( 5,4),( 5,5),( 5,6),(1,5),(2,5),(3,5), (4,5),( 6,5)
probability of getting getting 5 =11/36
Probability of not getting 5 =1-(11/36) =25/36
probability of getting. 5 exactly one time is =10/36

Answer 2

Answer: The Probability that 5 will not come up on either of them is 25/36 and the Probability that 5 will come up exactly one time is 5/18.

Step-by-step explanation:

It is given in the question that a dice is rolled twice .

If we throw two dices then the total possible outcomes are as follows:

$(1,1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)$

$(2,1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)$

$(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)$

$(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)$

$(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)$

$(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)$

Thus, the total no. of possible outcomes when the two dices are thrown is 36

(i) Let E1 be the event that 5 will not come up either of them

Total number of favorable outcomes (5 not coming up on either of them) :  

$(1, 1) (1, 2) (1, 3) (1, 4) (1, 6)$

$(2, 1) (2, 2) (2, 3) (2, 4) (2, 6)$

$(3, 1) (3, 2) (3, 3) (3, 4) (3, 6)$

$(4, 1) (4, 2) (4, 3) (4, 4) (4, 6)$

$(6, 1) (6, 2) (6, 3) (6, 4) (6, 6)$

Total no. of favorable outcomes = 25  

Hence,P(E1) = Number of favourable outcomes/total number of outcomes

P(E1) = 25/36

Hence, the Probability that  5 will not come up on either of them, P(E1) is 25/36

(ii) Let E2  be the event that 5 will come up exactly one time  

Total number of favorable outcomes (5 will come up exactly one time) : {(1, 5) (2, 5) (3, 5) (4, 5)  (5, 1) (5, 2) (5, 3) (5, 4) (5, 6) (6, 5)}

Total no. of favorable outcomes = 10

Hence, P(E2) = Number of favourable outcomes / total number of outcomes

P(E2) = 10/36 = 5/18

Hence, the Probability that 5 will come up exactly one time , P(E2) is 5/18

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