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 $\frac{3}{8}$

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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 ”

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ANS : Option B

# Easy Explanation 

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

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best answer , the most intuitive

Answer 1

ANS : Option B

 

# Easy Explanation @Sachin Mittal 1

Method 1 :

here are 6 possible outcomes for a die roll. Out of these 3 are even and 3 are odd.

prob (odd) =prob (even) = 3/6 = ½

 

Target : We Want Exactly 1 Odd No  

Required outcome : {(odd,even,even), (even,odd,even), (even,even,odd) }

 

P (odd,even,even) = P1= ½ * ½ * ½ =1/8

P (even,odd,even) = P2= ½ * ½ * ½ =1/8

P (even,even,odd) = P3= ½ * ½ * ½ =1/8

Prob = P1 + P2 + P3 

          = 1/8 * 1/8 *1/8 

          =  3/8

Prob = 3/8

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Method 2 :

We can use Binomial distribution:

n = 3 trails

prob ( of success.) = prob (of getting odd no.) = ½

P(X=1)  =  3C1 * (1/2)1 *(1/2)2 = 3/8

Prob = 3/8

 

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Answer 2

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

Answer 3

The probability of an odd outcome from a single roll of a die = 1/2

Probability of odd an outcome when a die is rolled 3 times = $(1/2)^{_{3}}$

Probability of getting atleast 1 odd outcome = 1 – (1/8) = 7/8

Probability of getting exactly 1 odd = 7/8 – 1/2 = 3/8

 

 

Answer 4

Let the outcome when rolling one die ===== { 1, 2, 3, 4, 5, 6 } ////////////     it consist of --       odd no: { 1, 3, 5 } so probability of getting an odd = 3/8    ////////////////////     even no : { 2, 4, 6 }, so probability of getting even also- 3/8.

 possible outcome that comes when rolled 3 times are = { odd,even,even} or { even,odd,even } or { even,even,odd }

                                                                                =  { 3/8 * 3/8 * 3/8 } + { 3/8 * 3/8 * 3/8 } + { 3/8 * 3/8 * 3/8 }

                                                                                 = 1/8 * 3

                                                                                = 3/8 ( got the answer ). so correct one is option B.