Probabily: Permutation of words in a sample with replacement











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Form the six letters $A, B, C, D, E, F$ three letters are chosen at random with replacement. What is the probability that either the word BAD or CAD can be formed from the chosen letters.



a) $1/216$



b) $3/216$



c) $6/216$



d) $12/216$



My approach:



Possible outcomes $= 6 times 6 times 6 = 216$.



Favourable Outcome $ = 2 times 3! = 12.$



Hence Probability $= 12/216$. I am getting d). So where am I making a mistake?



Reference: CSIR NET DEC 2015 Paper A Q.No. 50



Edit: As it turns out I was checking the wrong answer key. My solution is right.






probability







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  • 1




    Looks right ... are you sure the correct answer is c?
    – Bram28
    Nov 20 at 4:07










  • @Bram28 Yes, you can check for yourself, CSIR NET DEC 2015 Part A Question 50.
    – henceproved
    Nov 20 at 4:09










  • Well, I still say it is d!
    – Bram28
    Nov 20 at 4:14










  • @BRam28 Sorry for the confusion, I rechecked it, as it turns out I was checking the wrong answer key. Mine (as well as yours) solution is right.
    – henceproved
    Nov 21 at 2:52










  • Aha! I figured :) Thanks for letting me know.
    – Bram28
    Nov 21 at 3:39

















up vote
2
down vote

favorite












Form the six letters $A, B, C, D, E, F$ three letters are chosen at random with replacement. What is the probability that either the word BAD or CAD can be formed from the chosen letters.



a) $1/216$



b) $3/216$



c) $6/216$



d) $12/216$



My approach:



Possible outcomes $= 6 times 6 times 6 = 216$.



Favourable Outcome $ = 2 times 3! = 12.$



Hence Probability $= 12/216$. I am getting d). So where am I making a mistake?



Reference: CSIR NET DEC 2015 Paper A Q.No. 50



Edit: As it turns out I was checking the wrong answer key. My solution is right.






probability







share|cite|improve this question




















  • 1




    Looks right ... are you sure the correct answer is c?
    – Bram28
    Nov 20 at 4:07










  • @Bram28 Yes, you can check for yourself, CSIR NET DEC 2015 Part A Question 50.
    – henceproved
    Nov 20 at 4:09










  • Well, I still say it is d!
    – Bram28
    Nov 20 at 4:14










  • @BRam28 Sorry for the confusion, I rechecked it, as it turns out I was checking the wrong answer key. Mine (as well as yours) solution is right.
    – henceproved
    Nov 21 at 2:52










  • Aha! I figured :) Thanks for letting me know.
    – Bram28
    Nov 21 at 3:39















up vote
2
down vote

favorite









up vote
2
down vote

favorite











Form the six letters $A, B, C, D, E, F$ three letters are chosen at random with replacement. What is the probability that either the word BAD or CAD can be formed from the chosen letters.



a) $1/216$



b) $3/216$



c) $6/216$



d) $12/216$



My approach:



Possible outcomes $= 6 times 6 times 6 = 216$.



Favourable Outcome $ = 2 times 3! = 12.$



Hence Probability $= 12/216$. I am getting d). So where am I making a mistake?



Reference: CSIR NET DEC 2015 Paper A Q.No. 50



Edit: As it turns out I was checking the wrong answer key. My solution is right.






probability







share|cite|improve this question















Form the six letters $A, B, C, D, E, F$ three letters are chosen at random with replacement. What is the probability that either the word BAD or CAD can be formed from the chosen letters.



a) $1/216$



b) $3/216$



c) $6/216$



d) $12/216$



My approach:



Possible outcomes $= 6 times 6 times 6 = 216$.



Favourable Outcome $ = 2 times 3! = 12.$



Hence Probability $= 12/216$. I am getting d). So where am I making a mistake?



Reference: CSIR NET DEC 2015 Paper A Q.No. 50



Edit: As it turns out I was checking the wrong answer key. My solution is right.






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 15:46









N. F. Taussig

42.9k93254




42.9k93254










asked Nov 20 at 3:57









henceproved

1308




1308








  • 1




    Looks right ... are you sure the correct answer is c?
    – Bram28
    Nov 20 at 4:07










  • @Bram28 Yes, you can check for yourself, CSIR NET DEC 2015 Part A Question 50.
    – henceproved
    Nov 20 at 4:09










  • Well, I still say it is d!
    – Bram28
    Nov 20 at 4:14










  • @BRam28 Sorry for the confusion, I rechecked it, as it turns out I was checking the wrong answer key. Mine (as well as yours) solution is right.
    – henceproved
    Nov 21 at 2:52










  • Aha! I figured :) Thanks for letting me know.
    – Bram28
    Nov 21 at 3:39
















  • 1




    Looks right ... are you sure the correct answer is c?
    – Bram28
    Nov 20 at 4:07










  • @Bram28 Yes, you can check for yourself, CSIR NET DEC 2015 Part A Question 50.
    – henceproved
    Nov 20 at 4:09










  • Well, I still say it is d!
    – Bram28
    Nov 20 at 4:14










  • @BRam28 Sorry for the confusion, I rechecked it, as it turns out I was checking the wrong answer key. Mine (as well as yours) solution is right.
    – henceproved
    Nov 21 at 2:52










  • Aha! I figured :) Thanks for letting me know.
    – Bram28
    Nov 21 at 3:39










1




1




Looks right ... are you sure the correct answer is c?
– Bram28
Nov 20 at 4:07




Looks right ... are you sure the correct answer is c?
– Bram28
Nov 20 at 4:07












@Bram28 Yes, you can check for yourself, CSIR NET DEC 2015 Part A Question 50.
– henceproved
Nov 20 at 4:09




@Bram28 Yes, you can check for yourself, CSIR NET DEC 2015 Part A Question 50.
– henceproved
Nov 20 at 4:09












Well, I still say it is d!
– Bram28
Nov 20 at 4:14




Well, I still say it is d!
– Bram28
Nov 20 at 4:14












@BRam28 Sorry for the confusion, I rechecked it, as it turns out I was checking the wrong answer key. Mine (as well as yours) solution is right.
– henceproved
Nov 21 at 2:52




@BRam28 Sorry for the confusion, I rechecked it, as it turns out I was checking the wrong answer key. Mine (as well as yours) solution is right.
– henceproved
Nov 21 at 2:52












Aha! I figured :) Thanks for letting me know.
– Bram28
Nov 21 at 3:39






Aha! I figured :) Thanks for letting me know.
– Bram28
Nov 21 at 3:39

















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