A basketball player shoots free throws and makes them with probability 1/7. What is the probability of the event the player will miss three in a row?

216/343 or 0.6297 (4dp)

Step-by-step explanation:

Anytime the probability of an event occurs more than once in a particular order (one time after another in the same instance), then the probability is multiplied together. If the probability of him making the shot is 1/7, then the probability of him missing it is 6/7. This is because this is the only other possible outcomes and all outcome probabilities must add to 1 (1-1/7 - 6/7).

So, if he misses it three times in a row then the probability is:

6/7 (probability miss this first time) x 6/7 (probability miss second time) x 6/7 (probability miss 3rd time).

6/7 x 6/7 x 6/7 = 216/343

or 0.6297 (4dp)

Hope this helped!

2160ft3

step-by-step explanation: