Hypergeometric Distrib, how to change up from exactly to at least

[mfdoom11]

Suppose there are 10 items in a lot. 5 are defective and 5 are non-defective. 6 are inspected. Find the following, rounded to 4 decimals.

a) Find the probability that exactly 2 of those chosen are defective.

b) Find the probability that at least 2 of those chosen are defective.


Little stuck on the steps between the two and how they differ

 

[pka]

Suppose there are 10 items in a lot. 5 are defective and 5 are non-defective. 6 are inspected. Find the following, rounded to 4 decimals.
a) Find the probability that exactly 2 of those chosen are defective.
b) Find the probability that at least 2 of those chosen are defective.
Little stuck on the steps between the two and how they differ

Both of these are very simple combinations. We have five defected and five non-defected.
Selecting six from ten and asking if two are defected? Look at this calculation. Study that setup.
Then explain to yourself why the second is answered as
\(\displaystyle \left(\sum\limits_{k = 2}^6 \binom{5}{k}\binom{5}{6-k}\right) \div \binom{10}{6} \)

 

[Steven G]

Suppose there are 10 items in a lot. 5 are defective and 5 are non-defective. 6 are inspected. Find the following, rounded to 4 decimals.

a) Find the probability that exactly 2 of those chosen are defective.

b) Find the probability that at least 2 of those chosen are defective.


Little stuck on the steps between the two and how they differ

Exactly 2 defective, well it means exactly 2 are defective.
At least 2 defective means exactly 2 are defective OR exactly 3 are defective OR exactly 4 are defective OR exactly 5 are defective. Why did I stop at exactly 5 are defective? Could I have gone on to exactly 6 are defective up to exactly 12 are defective? Why did I capitalize OR?