[email protected] wrote:Hi tmontgomery,
What is the source of this question? I asked because the question is worded in an odd way, but I think the 'intent' of the question is the same as asking:
"There are 6 balls in a box - 2 red and 4 green. Balls will be randomly drawn one at a time and not replaced. What is the probability of selecting the 4 green balls on the first four draws?"
Under these conditions, we can deal with each individual 'event', then multiply the results.
Probability of drawing a green on the
first try: 4/6
second try: 3/5
third try: 2/4
fourth try: 1/3
(4/6)(3/5)(2/4)(1/3) = 1/15
Final Answer:
B
The idea works, but it'd be more like this:
Prob(first letter is T, R, A, or M) = 4/7
Prob(second letter is T, R, or A) = 3/6
Prob(third letter is T or R) = 2/5
Prob(fourth letter is T) = 1/4
Which gives (4/7) * (3/6) * (2/5) * (1/4), or (1/35).
