Determine the sample space for an experiment if a coin is selected from a bag three times. There are three coins in the bag: a nickel (N), a quarter (Q), and a dime (D). Each time a coin is selected, it is replaced before selecting the next coin. (1 point)

1. NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, QQQ, QQD, QQN,
QDQ, QDN, QDD, QNQ, QND, QNN

2. NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, DDD, DDQ, DDN,
O DQD, DQN, DQQ, DND, DNQ, DNN, QQQ, QQD, QQN, QDQ, QDN, QDD,
QNQ, QND, QNN

3. NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, NNN, DDQ, DDN,
O DQD, DQN, DQQ, DND, DNQ, DNN, DDD, QQD, QQN, QDQ, QDN, QDD,
QNQ, QND, QNN

4. NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, DDD, DDQ, DDN,
DQD, DQN, DQQ, DND, DNQ, DNN

1 answer

To determine the sample space for the experiment of selecting a coin from a bag containing a nickel (N), a quarter (Q), and a dime (D) three times (with replacement), we need to consider all the possible combinations of selections.

Since there are 3 coins and each selection is independent (due to the replacement), we can represent the sample space using ordered triples (where each position in the triple corresponds to one selection). Each position can be filled with either N, Q, or D.

The total number of possible outcomes is given by \(3^3\) (since there are 3 choices for the first selection, 3 for the second selection, and 3 for the third selection), resulting in \(27\) outcomes.

The sample space consists of all combinations:

  • For the first position, you can select either N, Q, or D.
  • The same options apply for the second and third positions.

Thus, the sample space includes the following combinations:

  1. NNN
  2. NNQ
  3. NND
  4. NQN
  5. NQD
  6. NQQ
  7. NDN
  8. NDQ
  9. NDD
  10. QNN
  11. QNQ
  12. QND
  13. QQN
  14. QQD
  15. QDQ
  16. QDN
  17. QQQ
  18. QDQ
  19. QDD
  20. DNN
  21. DNQ
  22. DND
  23. DDQ
  24. DQD
  25. DQN
  26. DQQ
  27. DDD

Regarding the options provided:

  • Option 1 and Option 2 both include incorrect strings and invalid combinations.
  • Option 3 has repeated elements and also invalid combinations.
  • Option 4 lists all the correct combinations without any repetitions or missing combinations.

Therefore, the correct answer is:

4. NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, DDD, DDQ, DDN, DQD, DQN, DQQ, DND, DNQ, DNN, QQQ, QQD, QQN, QDQ, QDN, QDD.