Not Sure about the fastest way to do this. But here it goes:
We take into cosndieration all possibilities from 1 par to all 4 pars.
1) Probability of Making 1 Par and missing the other is 0.3 x 0.7 x 0.7 x 0.7 i.e 0.1029 which can be arranged in 4 ways (4 x 0.1029).
2) Probability of Making 2 Pars and missing the other is 0.3 x 0.3 x 0.7 x 0.7 i.e 0.0041 which can be arranged in 6 ways (6 x 0.0041).
3) Probability of Making 3 Pars and missing the other is 0.3 x 0.3 x 0.3 x 0.7 i.e 0.0189 which can be arranged in 4 ways (4 x 0.0189).
4) Probability of Making 1 Par and missing the other is 0.3 x 0.3 x 0.3 x 0.3 i.e 0.0081 which can be arranged in 1 way (1 x 0.081).
The total comes out to 0.7599. Hence, the correct option is D.
I believe the faster method is to do 1 - zero successes but I am still getting better at it so the long way above.
Bunuel
When Matthew plays his favorite golf course, his probability of making a par or better on a par-3 hole is .3. If Matthew is playing his favorite golf course, and that course has exactly 4 par-3 holes, what is the probability that Matthew will make par or better on at least one of the par-3 holes?
A. 0.0081
B. 0.2401
C. 0.3000
D. 0.7599
E. 0.9919