Question

A die is thrown twice. What is the probability of getting 2 at most once?

Answer 1

As the die is thrown twice so ,

Total number of possible outcomes = 36

Now,


Total number of outcomes with a double 2s = 1
Probability of getting a double 2 = 1/36

P(Getting almost a 2) = 1 - P(getting both 2s)

P(Getting almost a 2) = 1 - 1/36

P(Getting almost a 2) = 35/36

Answer: 35/36

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Answer 2

Your answer goes like this . . .

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No. of outcomes when a die is thrown once = $6$

No. of outcomes when a die is thrown twice = $6^ {2}$

The sample space will be created as under :-

( 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)

Now, we can split the question into two parts viz (i) and (ii).

(i) There will be no '2' in any throw:

There will be 25 outcomes because the first throw may show any of the 5 numbers 1, 3, 4, 5 & 6 and for each of the outcome in the first throw, the second throw may show any of those 5 numbers viz 1,2,3,4 & 6. And its probability is $\frac{25}{36}$.

(ii) '5' will appear only in one throw:

There will be 10 outcomes viz (1,2),(3,2),(4,2),(5,2),(6,2),(2,1),(2,3),(2,4),(2,5) & (2,6). And its probability is $\frac{10}{36}$.

= = = = = =

Hence, the required probability will be :-

P (getting 2 at most once) = $\frac{25}{36}$ + $\frac{10}{36}$ = $\frac{35}{36}$.
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