Dhairya275 wrote:
Each of four different locks has a matching key. The keys are randomly reassigned to the locks. What is the probability that exactly two of the keys fit the locks to which they are reassigned?
1. 1/8
2. 1/6
3. 1/4
4. 3/8
5. 1/2
Help please ! Any Simple Solution ?
Dear
Dhairya275There are not hugely simple solutions to this. All solutions that occur to me involve using counting techniques. You might take a look at this blog post.
http://magoosh.com/gmat/2013/gmat-proba ... echniques/Total number of orders for 4 keys = 4! = 4*3*2*1 = 24
Of those 24 possible orders, how many have two keys in the right place and two in the wrong place. Suppose the locks are {a, b, c, d}. If the keys are in the order {A, B, C, D}, then all four are correct. To get two right & two wrong, we would need to select one pair from {A, B, C, D} and switch them. How many different pairs can we select from a set of four?
4C2 = \(\frac{4!}{(2!)(2!)}\)
= \(\frac{4*3*2*1}{(2*1)(2*1)}\)
= \(\frac{4*3}{2}\)
= 6
So, of the 24 sets, 6 of them would have two right & two wrong. P = 6/24 = 1/4
Does this make sense?
Mike
Thank you . It helped me a lot .