a die is thrown.

if even pick 1 mug from bag having 3 black and 4 white mugs.
if odd pick 2 mugs.

what is the probability that the second mug is white

5 answers

Let's first consider the probability of picking two white mugs from the given back.

Probability of first white mug = White/Total = 4/7

Probability of second white mug = White/New Total = 3/6 = 1/2

So, the probability of picking two white mugs in a row from that bag is (4/7)*(1/2) = 2/7

Now, note that the question asks for the probability that the *second* mug is white, and you only pull a second mug out if you roll an odd number.

So the probability we just found out will be multiplied by (1/2) since an odd number is only rolled half the time.

Actual probability = (2/7)*(1/2) = 1/7
O - odd, E - even, B - black mug , W - white mug

possible outcomes and assuming the mug is not returned if 2 are picked:

E B = (1/2)(3/7) = 3/14
E W = (1/2)(4/7) = 4/14
O WW <--prob--> (1/2)(4/7)(3/6) = 1/7
O WB = (1/2)(4/7)(3/6) = 1/7
O BW <--prob--> (1/2)(3/7)(4/6) = 1/7)
O BB = 1/14

prob(your event) = 1/7+1/7 = 2/7

(note that the sum of the probs of the 6 cases = 1
to clarify the question more
if the outcome of the die is odd, we will pick only one mug so we will not have a second mug at all
sorry
if the outcome of the die is even, we will pick only one mug so we will not have a second mug at all
70% of new employees take a learning lessons.
During first month of work they have probability of 0.04 to make mistakes while those who didn't take lessons (30%) have probability of 0.09.
Given that an employee didn't make mistakes in his first month, what is the probability that he had token lessons