What is the probability of getting $8-6$ (Win-lose), when you bet 14 times in a win or lose a game with $60%$ winning percentage?
According to the answer, it must be $ 17.4$%.
I tried to solve this using binomial distribution but my answer is not $ 17.4$%. Here is my solution:
$B(x;n,p) = B(8;14,0.6)$ = $14 \choose 8$$(0.6)^8(0.4)^6$ $= (3003)(0.6)^8(0.4)^6$ $= 0.20659760529408$
Here is the exact problem
"Let’s say, based on your betting system’s 60% winning percentage, you want to know Harvard's most likely record over the next 14 bets. 60% of 14 is 8.4, so our record should be 8-6. Using the binomial distribution calculator, we learn that even though 8-6 is the most likely record, it will actually only occur 17.4% of the time. Show that it will actually only occur 17.4% of the time."
Please help me solve this problem. Any comments or suggestions will be much appreciated.
As you probabily already did, the probability to get 8 successes on 14 independent draws is
$$\binom{14}{8}0.6^8 0.4^6\approx 20.7\%$$
Edit: reading the text, the probability that there is an error (more than one error) is getting higher and higher.
Here is the graph of your 60% winning game (with 14 independent trials)
As you can see, the most likely events are 2, with same 20.7% probability