Chapter 11 - Probability
11.2 ["Not" and "Or" Problems] A fair die is rolled "Not" (a) Probability of rolling a 3 or a 5 (b) An odd # or a # less than three "Or" (c) Not greater than 4
(a) P(3 or 5)= P(3)+P(5) - P(3 & 5) 1/6 + 1/6 - 0 = 2/6 = 1/3 (b) P(odd or x<3)= P(odd) + P(x<3) - P(odd & x<3) 3/6 + 2/6 - 1/6 = 4/6 = 2/3 (c) P(not 4<x) = 1 - (greater than 4) 1 - 2/6 or 6/6 - 2/6 = 4/6 =2/3
11.3 [Conditional Probability and "And" Problems pt.2] If you are dealt two cards successively (w/replacement on the first) from a standard 52-card deck, find the probability of getting: (a) a heart on the first card and a diamond on the second (b) a face card on the first card an ace on the second
(a) P(heart & diamond) = 13/52 * 13/52 = 1/16 (b) P(face & ace) 12/52 * 4/52= 3/169
11.3 [Conditional Probability and "And" Problems] "Conditional Probability" If two fair dice are rolled, find the probability that (a) the sum is 6 given that the roll is a double (b)the numbers rolled dorm a "double" given that their sum is 11 "And" (c) If a fair coin is tossed three times, find the probability of getting heads on the first toss and tails on the second and third tosses. (d) A family has five children. The probability of having a girl is 1/2. What is the probability of having three girls followed by two boys?
(a) P(sum of 6\double) = 1/6 (3,3) is the only double that can make 6 (b) P(double\sum of 11) = 0/2 There are no ways to receive doubles to create a sum of 11. There are two ways in which to get a sum of an 11,(5,6) and (6,5) (c) P(H&T&T) = 1/2 * 1/2 *1/2 = 1/8 (d) P(GGGBB) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32
11.1 [Basic Concepts] Coin Toss- Three fair coins are tossed. (a) Write out the sample space. (b) Determine the probability of no tails. (c) Determine the probability of exactly one tail. (d) Determine the probability of at least two tails. (e) Determine the probability of exactly three tails.
(a) Sample space: { hhh, thh, hth, hht, tth, tht, htt, ttt} (b) The probability of no tails is: {1/8} (c) The probability of exactly one tail is: {3/8} (d) The probability of at least two tails is: {1/2} (e) The probability of exactly three tails is: {1/8}
11.5 [Expected Value] Net winnings
(amount you win/earn)-(amount you paid)
11.4 [Binomial Probability] p-probability to success, q-probability of failure, x-successes(choose), n-total# of outcomes (nCx)(p^x)(q^n-x)=(n!/x![n-x!])(p^x)(q^n-x) A coin is tossed 5 times. Find the probability that (a) exactly 3 are heads (b) at least 3 are heads (nCx)(p^x)(1−p)^n−x [Note:1-p because q isn't given] A coin is tossed 5 times. Find the probability that (c) exactly 1 is a tail (d) none are tails/all are tails (e) at least 3 tails
Parts (a)-(b): p-0.5, q-0.5, x-3, n-5 (a)P(exactly 3 h)= (5C3)(0.5^3)(0.5^5-3=2) ≈0.3130 ≈0.313 (b) P(0,1,or 2 h)=P(0h)+P(1h)+P(2h)= repeat the steps from the last problem and add the solutions all together ≈ 0.5 Part (c): p-0.5, q-0.5, x-1, n-5 (5C1)(0.5^1)(1-0.5^4) ≈0.156 Part (d): x-5 (5C5)(0.5^0)(1-0.5^5) ≈0.031 Part (e): x-0,1,2,3 P(0)= P(1)= P(2)= P(3)= ≈0.500
11.4 [Binomial Probability pt.3] B(n,k;p)=C(n,k)(p^k)(1-p^n-k) Dr. Kenobi has developed a new procedure that they believes can correct a life-threatening medical condition. If the success rate for this procedure is 65% and the procedure is tried on 10patients, what is the probability that at least 7 of them will show improvement?
n-10, k-7, p-0.65, q- 0.35 C(10,7)= (10C8)(0.65^8)(0.35^10-7) =0.2522 C(10,8)= (10C8)(0.65^8)(0.35^10-8) =0.1757 C(10,9)= (10C9)(0.65^8)(0.35^10-9)=0.0725 C(10,10)= (10C10)(0.65^8)(0.35^10-10)=0.0135 Add the sum of those possibilities 0.2522+0.1757+0.0725+0.0135 =0.5139
11.4 [Binomial Probability pt.2] A fair die is rolled four times. A 6 is considered "success", while all other outcomes are "failures". Find the probability of 2 successes.
p-1/6 n-4 q-5/6 x-2 (4C2)(1/6^2)(5/6^4-2) =25/216
11.5 [Expected Value] A certain game consists of rolling a single fair die and pays off as follows: $5 for a 6, $3 for a 5, $2 for a 4, and no payoff otherwise. Find the expected winnings for this game.
x1, x2, x3, and x4 winnings are $7, $4, $1, and $0 Winnings x-$5 P(x) 1/6 =$0.833 Winnings x-$3 P(x) 1/6 =$0.500 Winnings x-$2 P(x) 1/6 =$0.333 Winnings x-$0 P(x) 3/6 =$0 Add the products to find the expected value =$1.67