GATE CSE 1998 | Question: 1.1

Question

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

• A. $\dfrac{1}{6}$  
• B. $\dfrac{3}{8}$  
• C. $\dfrac{1}{8}$  
• D. $\dfrac{1}{2}$

Comments

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 $\frac{3}{8}$

---

If we take Last has odd numbers we get, (2,2,1), (2,2,3), (2,2,5), (2,4,1), (2,4,3), (2,4,5), (2,6,1), (2,6,3), (2,6,5) = 9 , we repeat the same for 4 and 6 we get, 9*3 = 27 favorable outcomes. Total outcomes 6^3= 216.

Probability for last element as odd = 27/216 = 1/8

If we add for first and middle element as odd = 1/8 +1/8 + 1/8 = 3/8

Answer → “ B ”

---

ANS : Option B

# Easy Explanation 

https://gateoverflow.in/1638/gate-cse-1998-question-1-1?show=412900#a412900

---

best answer , the most intuitive

Answer 1

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 = $3/8$.

Comments

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

---

Nice!

---

@Angkit   

 – – –  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

Answer 2

We can use Binomial:

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

Answer 3

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

Answer 4

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