2. (i) is better, since the extra condition "and the second is a queen" in (ii) makes winning occur less often.
4. (4/52)(3/51)(2/50)(1/49)(4/48), or about 0.0000003.
5. Yes, color and number are independent. The probability of getting an 8 is 2/6 = 1/3. The probability of getting an 8 given that the ticket is white is still 1/3. The probabilities for numbers don't change when information about color is given (and vice versa).
6. (a) True, since there are 52 possible cards and 1 ace of
clubs.
6.(b) True, since we don't know what the first card is, the second card
still has 52 possible outcomes 1 of which is the ace of diamonds.
6.(c) False, P=(1/52)(1/51) since the events are not independent.
8. (a) P=(4/6)4 = 16/81, or about 0.20.
8.(b) P=(2/6)4 = 1/81, or about 0.01.
8.(c) Use the NOT rule: P=1-P(all showing 3 or more spots) = 1-(16/81)
= .80.
9. (a) P=(1/6)10 = 1/60466176=0.000000017.
9.(b) Use the NOT rule: P=1-(1/6)10 = 60466175/60466176=
0.999999983.
9.(c) P=(5/6)10 = 9765625/60466176 = 16.15%.
11. One way, perhaps the only way, is to have the first ticket be 1,1, the seventh be 3,2 and the eighth be 3,3. Then whether the first number is a 1 or a 3, the second number has a 1/4 chance of being a 1, a 1/2 chance of being a 2, and a 1/4 chance of being a 3.
1. (a) P = 1/36 = (1/6)(1/6). (b) P= 6/36 using the chart on page 239.
2. P = 2/36 = 1/18.
5. (a) False; (b) True, as long as the two events A and B have positive probability. Mutually exclusive means they cannot both happen, so the conditional probability P(A|B), 0, is not the same as P(A), and thus they are not independent.
6. "If you want to find the chance that at least one of the two events will happen, check to see if they are mutually exclusive; if so, you can add the chances." "If you want to find the chance that both events will happen, check to see if they are independent; if so, you can multiply the chances."
7. The key words here are: "at least once". We can simplify the problem by using the NOT rule. The probability that a 2 is never drawn is P= (3/5)(3/5)(3/5)(3/5)=0.1296, so the probability that it is drawn at least once is 1 - 0.1296 = 0.8704.
8. Four draws without replacement will leave only one tile left in the box. Even if that tile is a 2, the other 2 must have been drawn, so at least one 2 will ALWAYS be drawn: P=1.
9. There are a total of 12 outcomes: 3 from the first box and 4 from the second. We can count the number of ways A is larger than B, A is equal to B, and A is smaller than B. So: (a) P=3/12 = 1/4; (b) P=3/12 = 1/4; (c) P=6/12 = 1/2.
10. (ii) is better since you have a 3 out of 6 chance of winning each time you play, while with (i) you have only a 2 out of 6 chance of winning. (If the box had had four 0's and two 1's, the games would have been equivalent.)
2. (iii), since the probability that it lands on a non-six is 5/6 on each (independent) roll. [(i) is P(all sixes); (ii) is P(some non-sixes); (iv) is P(at least one six)].
3. There are 16 different equally likely orders of girls and boys. Of these, one is all girls; and for one boy, there are 4 (a boy in each of the 4 places). So, 5/16 of 4-child families should have more girls than boys.
4. False: Since the draws are made without replacement, the trials do not have the same probability of drawing green marbles. The binomial formula for probability does not apply.
5. True: The number of ways to choose two things from eight is (8*7)/(2*1) = 28, while the number of ways to choose 5 from 8 is (8*7*6*5*4)/(5*4*3*2*1) =(8*7*6)/(3*2*1) =56.
6. False: They are equal, since to choose 2 members to be on the committee is essentially to choose 6 members not to be on the committee. The number of ways is (8*7*6*5*4*3)/(6*5*4*3*2*1) = (8*7)/(2*1) = 28.
8. Let C(n,k) denote the number of ways of choosing k things from a group of n. Here, we have two events with five rolls each. The two events are independent. P(2 heads on 1st 5 rolls) = C(5,2)(1/2)2(1/2)3, and P(4 heads on 2nd 5 rolls)= C(5,4)(1/2)4(1/2). For them both to happen (AND rule) we multiply P= 10/(25) * 5/(25) = 50/(210) = 0.0488.
11. (a) We have 22 "trials", 17 "successes" and we are supposing that it is
an equal chance that the smoker or non-smoker dies first (p=1/2);
The binomial formula gives us the chance for EXACTLY 17 "successes".
We want the chance of 17 or more, so we need the probability of 17,18,19,20,21,
or 22 "successes". The NOT rule would make it even worse, so we chug through:
P = C(22,17) (1/2)17 (1/2)5
+ C(22,18) (1/2)18 (1/2)4
+ C(22,19) (1/2)19 (1/2)3
+ C(22,20) (1/2)20 (1/2)2
+ C(22,21) (1/2)21 (1/2)1
+ C(22,22) (1/2)22 (1/2)0
= (26334 + 7315 + 1540 + 231 +22 +1)/(222) = 0.00845.
There is a .845% probability that 17 or more smoker deaths could have
happened first just by chance.
11. (b) Here, we have 9 "trials" all of which were "successes". So the
probability that they happened by chance is
P = C(9,9) (1/2)9 = 1/512.
11. (c) Here, we have 2 "trials" with two "successes". So the
probability of this happening by chance is P=C(2,2)(1/2)2 = 1/4.
11. (d) It is possible that they happened by chance, but unlikely. These
probablities are pretty low. The difference can't have been caused by genetics
because each pair was genetically identical. The health effects of smoking
are a likely candidate for the cause of the difference, but one study is
not 100% proof.