Math, asked by mericabastola, 6 months ago

two dice are rolled two times find the probability of not getting 5 in both drawing​

Answers

Answered by vidhikagarg2007
1

Answer:

2/2 = 1

Step-by-step explanation:

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Answered by srijit64
2

Answer:

ANSWER

In a throw of pair of dice, total no of possible outcomes=36(6×6) which 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)

Solution(i):

Let E be event of getting 5 on either time

No. of favorable outcomes =11 (i.e.,(1,5)(2,5)(3,5)(4,5)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,5))

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

36

11

We have, P(E)+P(

E

ˉ

)=1

P(

E

ˉ

)=1−P(E)=1−

36

11

=

36

25

Therefore, the probability of not getting 5 either time =

36

25

Solution(i):

Let E be event of getting 5 exactly one time

No. of favorable outcomes=10 (i.e.,(1,5)(2,5)(3,5)(4,5)(5,1)(5,2)(5,3)(5,4)(5,6)(6,5))

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

36

10

=

18

5

Therefore, the probability of getting 5 exactly one time =

18

5

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