A box contains 11 red chips and 4 blue chips. We perform the following two-step experiment: (1) First, a chip is selected at random from the box and is then removed from the box. (After this first step, there are 14 chips left in the box. ) (2) Then, a chip is selected at random from the box (that is, from the remaining 14 chips) . Let B 1 be the event that the chip removed from the box at the first step of the experi- ment is red. Let B 2 be the event that the chip removed from the box at the first step of the experiment is blue. Let A be the event that the chip selected from the box at the second step of the experiment is red. Find P(B1), P(B2), P(A), P(B1|A), and P(B2|A).

P(B1) = (11/15)

P(B2) = (4/15)

P(A) = (11/15)

P(B1|A) = (5/7)

P(B2|A) = (2/7)

Step-by-step explanation:

There are 11 red chips and 4 blue chips in a box. Two chips are selected one after the other at random and without replacement from the box.

B1 is the event that the chip removed from the box at the first step of the experiment is red.

B2 is the event that the chip removed from the box at the first step of the experiment is blue. A is the event that the chip selected from the box at the second step of the experiment is red.

Note that the probability of an event is the number of elements in that event divided by the Total number of elements in the sample space.

P(E) = n(E) ÷ n(S)

P(B1) = probability that the first chip selected is a red chip = (11/15)

P(B2) = probability that the first chip selected is a blue chip = (4/15)

P(A) = probability that the second chip selected is a red chip

P(A) = P(B1 n A) + P(B2 n A) (Since events B1 and B2 are mutually exclusive)

P(B1 n A) = (11/15) × (10/14) = (11/21)

P(B2 n A) = (4/15) × (11/14) = (22/105)

P(A) = (11/21) + (22/105) = (77/105) = (11/15)

P(B1|A) = probability that the first chip selected is a red chip given that the second chip selected is a red chip

The conditional probability, P(X|Y) is given mathematically as

P(X|Y) = P(X n Y) ÷ P(Y)

So, P(B1|A) = P(B1 n A) ÷ P(A)

P(B1 n A) = (11/15) × (10/14) = (11/21)

P(A) = (11/15)

P(B1|A) = (11/21) ÷ (11/15) = (15/21) = (5/7)

P(B2|A) = probability that the first chip selected is a blue chip given that the second chip selected is a red chip

P(B2|A) = P(B2 n A) ÷ P(A)

P(B2 n A) = (4/15) × (11/14) = (22/105)

P(A) = (11/15)

P(B2|A) = (22/105) ÷ (11/15) = (2/7)

Hope this Helps!!!

step-by-step explanation:

see if this :

2 over 5

step-by-step explanation:

the 2: 3 ratio is the length of the ac segment to the cb segment total length. this implies that the ratio between segment ac length and segment ab length is 2: 5. point c is 2/5 from a to b, so the required fraction is 2/5.