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

Possible outcomes = {(0,1,1),(1,0,1),(1,1,0)} = 3

Sample space 2^{3 }= 8

Hence $\frac{3}{8}$

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33 votes

Best answer

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 = $3/8$.

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 = $3/8$.

– – – 3 places , 3 odd numbers (1,3,5) , 3 even numbers (2,4,6)

Now fixed **first** place containing odd number (we can have exactly one odd), so for first place, we will have 3 choices (1,3,5), for the second place we will have 3 choices (2,4,6), for the third place we will have 3 choices (2,4,6). so 3*3*3 =3$^{3}$ ways.

Now fixed **second** place containing odd number (we can have exactly one odd), so for second place, we will have 3 choices (1,3,5), for the first place we will have 3 choices (2,4,6), for the third place we will have 3 choices (2,4,6). so 3*3*3 =3$^{3}$ ways.

Now fixed **third** place containing odd number (we can have exactly one odd), so for third place, we will have 3 choices (1,3,5), for the second place we will have 3 choices (2,4,6), for the first place we will have 3 choices (2,4,6). so 3*3*3 =3$^{3}$ ways

These are favourable outcomes, now total outcomes will be, we will have 6 choices (1,2,3,4,5,6) for each of the 3 places, so 6*6*6 = 6$^{3}$ ways

P = favourable outcomes / total outcomes =( 3*3$^{3}$) / 6$^{3}$ = 3/8

1

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