Question 1191996
<br>
The probability of his clearing is 0.6, so on any one jump the probability that he clears is 0.6 and the probability that he fails is 0.4.<br>
So on any given sequence of 7 jumps, the probability that he clears 3 times and fails 4 times is (0.6^3)*(0.7^4).<br>
The number of different sequences of 7 jumps in which he clears 3 times is "7 choose 3", C(7,3).<br>
ANSWER: C(7,3)*(0.6^3)*(0.7^4)<br>
You can do the calculations....<br>