Section 6-5

Note: Problems 2--16 were not assigned. (I was going to, but I changed my mind.)

(2) P(C) = 0.2 = 20%
(4) P(E) = 0.8 = 80%
(6) P(A ∩ E ) = 0.4 = 40%
(8) P(C ∩ E ) = 0.14 = 14%
(10) By formula for conditional probability P(A|E) = P(A ∩ E )/P(E) = 0.4/0.8 = 0.5 = 50%
(12) By formula for conditional probability P(C|E) = P(C ∩ E )/P(E) = 0.14/0.8 = 0.175 = 17.5%
(14) Are A and E independent? Yes. According to problem 10 above, P(A|E) = 50% = P(A).
(16) Are C and E independent? No. According to problem 12 above, P(C|E) = 17.4%, while the chart says P(C)=20%. In other words, P(C) drops from 20% to 17.4% if E happens.


(18) A fair die is rolled 5 times.
(A) What is the probability of getting a 6 on the 5th roll, given that the other rolls were also sixes?
The probability is 1/6. The rolls are independent.

(B) What is the probability that the same number comes up every time?
Think of an outcome as a sequence of 5 numbers between 1 and 6. For example, 21336 means first roll was 2, seconnd was 1, third and fourth were 3 and the fifth was 6. Then by the multiplication principle, there are 65 outcomes in the sample space S. Let E be the event "same number comes up each time."

Thus E = { 11111, 22222, 33333, 44444, 55555, 66666 }.

Therefore P(E) = n(E) / n(S) = 6 / 65 = 1 / 64 = 1 / 1296 ~ 0.000771604 ~ 0.077%.


(20) E = "pointer lands on odd number." F = "pointer lands on prime number."
That is E = {1, 3, 5} and F = {2, 3, 5}

(A) P(F|E) = P(F∩ E)/P(E) = P({3 ,5}) / P({1, 3, 5}) = 0.3 / 0.6 = 0.5 = 50%

(B) Are E and F independent? Well, we have P(F|E) = 50% from above, but the table shows P(F) = 40%. Consequently, P and F are dependent.


(22)
(A) P(N∩ R) = (0.7)(0.2) = 0.14 = 14%
(B) P(S) = (0.3)(0.6) + (0.7)(0.8) = 0.18 + 0.56 = 0.74 = 74%

 

 


(24) A cion is tossed twice, so S = {TT, TH, HT, HH}. Consider the events
E1 = {HT, HH} = "Head on first toss."
E3 = {TT, TH} = "Tail on second toss."
E4 = {HH, HT} = "Head on second toss."

(A) Since E1 ∩ E3 = {HT} ≠ ∅, it follows that these two events are not mutually exclusive.
Also, if one of them happens, the the probability of the other does not change. Thus they are independent.

(B) Since E3 ∩ E4 = ∅, these two events are mutually exclusive.
Also P( E3 ) = 50%, while P( E3 | E4 ) = 0% (it's the probability that there is a tail second given that there is a head second). It follows that these two events are dependent.


(28) Two cards are drawn from a 52-card deck. What's the probability that both cards are red given...

(A) the cards are drawn without replacement.
Let E be the event "first card is red," and let F be the event "second card is red."
We are looking to find P(E ∩ F ).
Now, P(E ∩ F ) = P(E)P(F|E) = (26/52)(25/51) = 650 / 2652 ~ 0.2451 ~ 24.51%

Note that you could also solve this as C26, 2 / C52, 2 ~ 24.51%.

(B) the cards are drawn with replacement.
Now the events E and F are independent, and P(E ∩ F ) = P(E)P(F) = (26/52)(26/52) = 1/4 = 25%


(30) A card is drawn from a 52-card deck. Consider the events
M = "the card is a diamond,"
N = "the card is an even number."

(A) Find P(N|M). Suppose M has happened. The drawn card could be any one of 13 diamonds.
Now, if N also happens, then the card could be any one of the 5 even diamond cards. Thus P(N|M) = 5/13.

(B) Are M and N independent? Yes. Because P(N) = 20/52 = 5/13, and this doesn't change if M has happened.


(40) A box contains 2 red, 3 white and 4 green balls. Two balls are taken out in succession. What is the probability that no white balls are drawn?

Let E be the event "The first ball is not white." Let F be the event "The second ball is not white." We seek P( E and F). Now, P(E and F) = P(E ∩ F) = P(E)P(F|E) = (6/9)(5/8) = 5/12 ~ 0.4166 ~ 41.66%.


(56) 
                      
                         /Voted
                        /
                    .6 /
                      /
                     /
                    /
            Democrat
           /        \
      .55 /          \
         /            \
        /           .4 \
       /                \Didn't vote
      /
      \                     Voted
       \                   /
        \                 /
     .45 \               /
          \             /
           \           /
            Not Democrat
                       \
                        \
                         \
                          \
                           Didn't vote
The tree for this problem is sketched above. Notice that there's not enough information to fill in all the probabilities, though there is enough information to answer the question. The probability that a random person is a Democrat and voted in the last election is (.55)(.6) = .33 = 33%