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A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

 (i) 25/36

(ii)11/36

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 HI .......

 i) 5/6

ii) 1/6

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bro  its nt 2 dice... 1ly 1....

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its thrown twice than

total no. of outcomes = 36

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mine is correct...............:P:P:P

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 Total number of outcomes = 6 × 6

= 36

(i)Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)

Hence, total number of favourable cases = 11

P (5 will come up either time) 

P (5 will not come up either time) 

(ii)Total number of cases, when 5 can come at least once = 11

P (5 will come at least once) 

  • 39

hey all plz make outcomes.....

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 got through ncert exercise15.1 Q24

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bt  they r nt thrwm simontaneously.....  its aftr othr

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 go through ncert exercise15.1 Q24

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samplespace=

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

(i) 5 will not come up either time?

ans)  p(e)= 25/36

(ii) 5 will come up at least once?

ans) 11/36

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1) 25/36

2)11/36

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the total no. of events is 12
i) 10/12 = 5/6
ii) 2/12 =1/6
it can't be 36 no. of outcomes because they are not saying find the probability of the products of the dice.
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as the die is rolled twice
6*6=total number of outcomes
=36
now favourable outcome-(5,1) (5,2) (5,3) (5,4) (5,5 )(5,6) and
(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
=11 outcomes.
so p(E)=favourable outcome /total outcome
p(E)=11/36
p(not E)=1-11/36
=36-11/36=25/36
probability of 5 not coming = 25/36

favourable outcomes=11
total outcome=36
p(E)=11/36
thus probability of 5 coming once=25/36

 
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8 April bad bataunga.tab tak wait karo. Expert answer
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yes..
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2/7
1/12
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yo
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What is the probability That the number 5 will not come Up in single throw Of A die
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solution is i not know.
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as the die is rolled twice
6*6=total number of outcomes
=36
now favourable outcome-(5,1) (5,2) (5,3) (5,4) (5,5 )(5,6) and
(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
=11 outcomes.
so p(E)=favourable outcome /total outcome
p(E)=11/36
p(not E)=1-11/36
=36-11/36=25/36
probability of 5 not coming = 25/36

favourable outcomes=11
total outcome=36
p(E)=11/36
thus probability of 5 coming once=25/36

?
  • 2
as the dice roll twice 6*6 =total no of out comes= 6*6=36
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2. P[ 5 atleast]. = 3/12=1/4
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