Can someone please explain how to do this on a TI-83 Calculator?

Question

A high school baseball player has a 0.262 batting average. In one game, he gets 6 at bats. What is the probability he will get at least 3 hits in the game?

Answer 1

The probability of getting a hit is 0.262 so the probability of not getting a hit is 0.738

The set of possible outcomes includes:

Hit, Hit, Hit, Hit, Hit, Hit

Hit, Hit, Hit, Hit, Hit, No Hit

. . . . . .

all the way down to

No Hit, No Hit, No Hit, No Hit, No Hit, No Hit,

One of these has 6 hits, some have 5 hits, some 4, and some 3. All of which count in our problem.

P(6 hits) = (0.262)⁶

P(5 hits) = (0.262)⁵(0.738)

P(4 hits) = (0.262)⁴(0.738)²

P(3 hits) = (0.262)³(0.738)³

To determine the number of possible outcomes that yield 6 hits, we use ₆C₆ on the TI-83 (the number of combinations of 6 items taken 6 at a time). To find out the number of possible outcomes that yield 5 hits, we use ₆C₅. And for 4 hits, ₆C₄, and lastly ₆C₃ for 3 hits.

So the total probability of getting 3 or more hits in a game we can do:

(₆C₆)(0.262)⁶ + (₆C₅)(0.262)⁵(0.738) + (₆C₄)(0.262)⁴(0.738)² + (₆C₃)(0.262)³(0.738)³

This calculation is probably really tricky on a TI-83 but I think it's possible in a one-line calculation. It just might not be worth it. The combinations are cumbersome requiring you to enter "6" then hit the math button and scroll to PROB then down to select nCr then hit the "6" all to do ₆C₆ and I'm sure you'd be best to put parenthesis around that.

I did it in a one line calculation in my TI-84 but it's quite a bit easier'

My answer was 0.18886 or 18.9% probability