a) Of 14 eggs in a refrigerator, 4 are bad. From these, 14 eggs 2 eggs are chosen at random to make a cake. What is the probability that (i) Both eggs are good. (ii) First is bad, second is good. (iii) First is good, second is bad. (iv) Both eggs are bad. Find the probability by with replacement as well as without replacement
Answers
Answer:
To get exactly 1 bad egg, obviously the other 3 have to be “not bad”.
We know from the question that there are 12 eggs of which 10 are “not bad” eggs and 2 are bad eggs.
Probability to get the 1 bad egg from the first pick is 2/12.
Probability to get 3 “not bad” eggs from the remaining picks is 10/11 x 9/10 x 8/9 = 8/11
Combine them and we get : 2/12 x 8/11 = 4/33
That is the probability of getting exactly 1 bad egg if the bad egg is picked first. We need to calculate the probability of the bad egg getting picked second, third, etc.
To calculate that you simply multiply it by 4C1 because there is 1 bad egg out of 4 picks (if there are 2 bad eggs then 4C2).
And you get : 4/33 x 4 = 16/33
Another way to solve it is :
You calculate how many ways can you pick 4 eggs out of 12. That would be 12C4. Then you calculate how many ways can you pick exactly 1 bad egg and 3 “not bad” eggs out of 12. That would be 2C1 x 10C3.
Then you make :
(2C1 x 10C3)/12C4
(2 x 120)/495
16/33
Same result.
Step-by-step explanation:
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