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I have solved in this way....

probabilty of geting atleast 3..

suppose probability of geting 1 and 2..is ..1/6+1/6..2/6=1/3..

so probability of geting atleast 3 is...1-probability of geting 1 and 2..we l get probabilty of geting atleast 3..rite?

so 1-1/3=2/3..
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yes this approach is also rit

The probability of each outcome is 1/6. The probability of getting at least a "3" is:

1/6+1/6+ 1/6 +1/6 = 2/3
or what you mentioned

probability of not getting atleast 3

1- (1/6 + 1/6) = 2/3

both methods are fine
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i approached like this ,
not getting a 3=5/6

atleast=1-5\6
= 1/6

pls explain what is wrong in the approach??
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yamshi06
i approached like this ,
not getting a 3=5/6

atleast=1-5\6
= 1/6

pls explain what is wrong in the approach??


Probability = Favorable Outcomes / Total Outcomes

A Die has 6 faces numbered from 1 through 6

i.e. Total Out comes = 6

Favorable outcomes = Numbers atleast 3 = numbers greater than or equal to 3 i.e. {3, 4, 5, 6} = 4

i.e. Probability = 4/6 = 2/3

ALTERNATIVELY

Probability = 1 -(Unfavorable Outcomes / Total Outcomes

Unfavorable outcomes = Numbers less than 3 = i.e. {1, 2} = 2

i.e. Probability = 1 -(2/6) = 1-(1/3) = 2/3

I hope it clears your doubt! :)
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yamshi06
i approached like this ,
not getting a 3=5/6

atleast=1-5\6
= 1/6

pls explain what is wrong in the approach??


Please Note "If you subtract the probability of NOT getting 3 from 1 then you will get Probability of getting 3"

whereas the question demand the probability of getting ATLEAST3
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we mainly do like dis,
P(at least one success) = 1 – P(no success)

i missing something in this concept plsd explain??
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yamshi06
we mainly do like dis,
P(at least one success) = 1 – P(no success)

i missing something in this concept plsd explain??

You are getting it Right in this context Yashmi

P(at least one success) = 1 – P(no success) is CORRECT

But Probability of buying atleast 2 apples will NOT be same as (1-probability of buying NO apple)

You must Understand that Probability of buying atleast 2 apples will be equal to (1-probability of NOT buying 2 or more apple)
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Given that Jessica rolls a die and We need to find what is the probability of getting at least a "3"?

As we are rolling one dice => Number of cases = \(6^1\) = 6

At least a 3 means 3 or more than 3
=> Possible cases are 3, 4, 5, 6 => 4 cases

=> P(Getting at least a 3) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

So, Answer will be \(\frac{2}{3}\)
Hope it helps!

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Watch the following video to MASTER Dice Rolling Probability Problems

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