I'm having a difficult time comprehending this solution and if someone can explain it to me I would very much appreciate it!
(this is not a OG question but a practice problem from MGMAT)
Question: For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?
Solution:
Step 1) (1/6)^3(5/6) + (1/6)^3(5/6)+(1/6)^3(5/6)+(1/6)^3(5/6) = 4(1/6)^3(5/6) --> probability of rolling a two exactly 3 times out of 4 attempts.
Step 2) ADD the probability of rolling a two exactly 4 times (1/6)^4
Final answer: 4(1/6)^3(5/6)+(1/6)^4
What I don't get is why do we have to add the probability of rolling two exactly 4 times to step 1? Doesn't step 1 answer the question which is the probability of rolling a two exactly 3 times out of 4? What am I missing here? :roll:
Thanks in advance for your help.
(this is not a OG question but a practice problem from MGMAT)
Question: For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?
Solution:
Step 1) (1/6)^3(5/6) + (1/6)^3(5/6)+(1/6)^3(5/6)+(1/6)^3(5/6) = 4(1/6)^3(5/6) --> probability of rolling a two exactly 3 times out of 4 attempts.
Step 2) ADD the probability of rolling a two exactly 4 times (1/6)^4
Final answer: 4(1/6)^3(5/6)+(1/6)^4
What I don't get is why do we have to add the probability of rolling two exactly 4 times to step 1? Doesn't step 1 answer the question which is the probability of rolling a two exactly 3 times out of 4? What am I missing here? :roll:
Thanks in advance for your help.
