This is fairly challenging question since there are number of "acceptable" or "successful" outcomes - OOBB, OBOB, OBBO, BOBO, BBOO, BOOB.
To appreciate this complexity let us compare it to a simpler question: What is the probability of all four dice landing on Blue or what is P(BBBB)?
Well, now we only have one "acceptable" or "successful" outcome: BBBB
To calculate the probability of any event we should split it into its parts or smaller events:
In this case, we have four events - throw of dice 1, 2, 3 and 4 - making up the overall throw of four dice.
The beauty of having only one desired outcome is that I know exactly what I want to get on each of the four throws and therefore can easily find the probability of desired outcomes for each throw:
----------Dice 1 Dice 2 Dice 3 Dice 4 Overall outcome
outcome --B------B-------B--------B-------- BBBB
Probab. ---1/6---1/6-----1/6------1/6-----(1/6)*(1/6)*(1/6)*(1/6) =
1 / 1296
Now, let us go back to the original question - what happens when we have more than one "good" outcome?
Again, "acceptable" or "successful" outcomes are OOBB, OBOB, OBBO, BOBO, BBOO and BOOB. I could rephrase by saying that I would be successful if I get OOBB
OR OBOB
OR OBBO
OR BOBO
OR BBOO
OR BOOB.
When in this situation we find the probability for each "successful" outcome and
ADD them (you might recall from the high school math that "OR" translates into "ADDITION"):
P(OOBB) + P(OBOB) + P(OBBO) + P(BOBO) + P(BBOO) + P(BOOB)
Each one of those six probabilities list above could be calculated exactly the same way we found P(BBBB), using the table above. However calculating six different probabilities would take really long and is impossible to manage within a reasonable timeframe.
Luckily, we need to find the probability for only one of those six outcomes:
P (OOBB) = 1 / 1296
Each one of the remaining five outcomes has exactly the same probability since all the outcomes are identical - two Bs and two Os - but in different order.
Thus:
Answer:
P(OOBB) + P(OBOB) + P(OBBO) + P(BOBO) + P(BBOO) + P(BOOB) = (1 / 1296) + (1 / 1296) + (1 / 1296) +
+ (1 / 1296) + (1 / 1296) + (1 / 1296) = 6* (1 / 1296) =
1 / 216
TAKE AWAYs:
1. Recognize the type of the question being tested:
Finding the probability of the event the has number of "acceptable" or "successful" outcomes. Strategy: Find the find the probability for each "successful" outcome and ADD them.
2. To calculate the probability of any event we should split it into its parts or smaller events.
Example: Instead of thinking of the throw of four dice, think of the four consecutive throws of one dice.
3. Identical outcomes acquired in a different order still have the same probability:
P(OOBB) = P(OBOB)
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