This question involves lots of calculation, which requires carefulness but is feasible in "4-5 mins".
Step 1: To reduce the chance of making calculation mistakes, I draw a table of probability before jumping into the calculation. (this is probably the most important step in solving the question)
First throw: Succeed 60% = 3/5 - Fail 40% = 2/5
After 1st throw:
If previous throw:
a. Succeed, then this current throw has chance: Succeed 80% = 4/5 - Fail 20% = 1/5
b. Fail, then this current throw has chance: Succeed 40% = 2/5 - Fail 60% = 3/5
* I convert everything to have 5 as denominator as the options provide hints of having 5^5 in denominator (converting saving time to reducing everything from 10 to 5)
Step 2: Calculation:
There are 5 ways the person has at least 4 successful free throws: 4 ways of failing ONLY 1 at any of the 5 throws AND 1 way of succeeding in all 5 throws.
Since each throw is influenced by previous throw result and influence next throw result, calculating 1 way and multiplying by 4 is a WRONG way to calculate the probability of having exactly 4 successful throws.
SSSSF: Prob = (3/5)*[(4/5)^3]*(1/5)
SSSFS: Prob = (3/5)*[(4/5)^2]*(1/5)*(2/5)
SSFSS: Prob = (3/5)*(4/5)*(1/5)*(2/5)*(4/5)
SFSSS: Prob = (3/5)*(1/5)*(2/5)*[(4/5)^2]
FSSSS: Prob = (2/5)*(2/5)*[(4/5)^3]
SSSSS: Prob = (3/5)*[(4/5)^4]
Since all options have 5^5 as dominators and my calculation is on the track, I calculate only the numerator.
Numerator = 3*(4^3) + 3* (4^2)*2 + 3*(4^2)*2 + 3*2*(4^2) + (2^2)*(4^3) + 3*(4^4)
= (4^2)*(12+6+6+6+16+48)
= 16* 94 = 1504
Long calculation, but fulfilling when the answer is correct. During the exam, I may take max 3-4 mins to calculate and probably skip if my result doesn't match any of the options.
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