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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. $\dfrac{1}{2}$
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Best answer
38 votes
38 votes
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 = $3/8$.
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27 votes
27 votes
We can use Binomial:

P(X=1) = 3C1 * (1/2)^1 *(1/2)^2 = 3/8
5 votes
5 votes
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
3 votes
3 votes

Let A = Event that exactly one odd number turns up

So,

$P(A) = \frac{n(A)}{n(S)} $

We have rolled a die 3 times. Assuming the rolls are independent, the outcomes will be in the form 

( _ )( _ )( _ )

So, for the sample space there are $(6).(6).(6) = 6^3$ outcomes.

Now, when we roll a die, the outcomes are $\{ 1,3,5\}$ and $\{2, 4, 6 \}$

For event A, we want exactly one odd number. So, the resultant outcomes will be in the form 

[odd](even)(even) OR (even)[odd](even) OR (even)(even)[odd]

So, for each of the above there are 3 ways to choose a odd number from $\{ 1, 3,5\}$ AND 3 ways to choose an even number from $\{2,4,6\}$ AND 3 ways to choose an even number from $\{2,4,6\}$, and hence the total number of outcomes for event A will be 

$[3]\cdot (3) \cdot (3) + (3) \cdot [3] \cdot (3) + (3) \cdot (3) \cdot [3] = 3 \cdot 3^3$

because, AND means multiplication and OR means addition

So, $P(A) = \frac{3\cdot 3^3}{6^3} = 3 \cdot (\frac{3}{6})^3 = 3 \cdot (\frac{1}{2})^3 = \frac{3}{8}$ 

Answer:

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