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A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is

1. $\dfrac{1}{6}$
2. $\dfrac{3}{8}$
3. $\dfrac{1}{8}$
4. $\dfrac{1}{2}$
edited | 2.2k views
+2

Let Odd $\rightarrow 0$ and Even $\rightarrow 1$

Possible outcomes = {(0,1,1),(1,0,1),(1,1,0)} = 3
Sample space 23 = 8

Hence $\frac{3}{8}$

Answer - $B$

There are $6$ possible outcomes for a die roll. Out of these $3$ are even and $3$ are odd. So, when we consider odd/even a die roll has only $2$ possible outcomes. So, for three rolls of the die we have $8$ possible outcomes.

Out of them only $3$ will have exactly one odd number$\left \{ OEE, EOE, EEO \right \}$

Probability = $3/8$.
edited
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Please tell me how you can get 8 possible solutions
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clear now?
+10
one possible approach-

$\frac{3^3+3^3+3^3}{6^3}=\frac{3\times3^3}{6^3}=3\times\left ( \frac{3}{6} \right )^3 =\frac{3}{8}$
0

plz explain in a bit more detail Sachin Mittal 1 ..

0
Nice!
We can use Binomial:

P(X=1) = 3C1 * (1/2)^1 *(1/2)^2 = 3/8
+1 vote
another method
We can have exactly one odd number.
We can get (odd,even,even ) + (even,odd,even) + (even,even,odd)
Favourable case = 3x3x3+3x3x3+3x3x3
Total sample space=216
Probability=3(3x3x3)/216
which gives 3/8

a die can have 1,2,3,4,5,6 in which 1,3,5 are odd and 2,4,6 are even

the odd number can turn up out of any three turns=3C1

the odd number selected =3C1/6C1

the even number selected =3C1/6C1

the even number selected =3C1/6C1

so,3C1* 3C1/6C1 * 3C1/6C1  * 3C1/6C1 =3/8

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