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… believe that method should be right, but do check for calculation errors ...
for the last part of the question, can treat as binomial because …
- the 5 trials are independent
- can treat as two possible outcomes, getting a -1, and not getting a -1.
- probability of getting a -1 remains constant for all the trials.
for the last part of the question, can treat as binomial because …
- the 5 trials are independent
- can treat as two possible outcomes, getting a -1, and not getting a -1.
- probability of getting a -1 remains constant for all the trials.