Total points 22

Section 6.3 was not included on the assignment

• Section 6.1 -- pp. 398 - 399
   
  • 2 (1 pt)

    What is the probability that a die comes up six when rolled?
     

    • 1/6
       
  • 4 (1 pt)

    What is the probability that a randomly selected day of the year is in April?
     

    • 30/366
       
  • 6   (2 pts)

    What is the probability that a card selected from a deck is an ace or a heart?

    E₁ = {AH, AC, AD, AS}

    E₂ = {AH, 2H, 3H, 4H, 5H, 6H, 7H, 8H, 9H, 10H, JH, QH, KH}

    E₁ Ç E₂ = AH

    |S| = 52

    p( E₁ U E₂ ) = p(E₁) + p(E₂) - p(E₁ Ç E₂)

    = 4/52 + 13/52 - 1/52

    = 16/52

  • 24 (4 pts)

    Find the probability of winning the lottery by selecting the correct six integers, where the order in which these integers are selected does not matter, from the positive integers not exceeding:
     

    1. 1/C(30, 6) = 1/593775
        
    2. 1/C(36, 6) = 1/1947792
        
    3. 1/C(42, 6) = 1/5245786
        
    4. 1/C(48, 6) = 1/12271512
        
• Section 6.2 -- pp. 414 - 415
   
  • 2 (1 pt)

    Find the probability of each outcome when a loaded die is rolled, if a 3 is twice as likely to appear as each of the other five numbers on the die.

    p(3) = 2p(s) when s ¹ 3

    p(1) = p(2) = p(4) = p(5) = p(6)
     

    p(S) = p(1) + p(2) + p(4) + p(5) + p(6) + p(3)

    = 5p(s) + 2p(s) when s ¹ 3

    = 7p(s) = 1

    p(s) = 1/7 when s ¹ 3

    p(3) = 2/7
     

  • 6 (3 pts)

    What is the probability of these events when we randomly select a permutation of {1, 2, 3}?
     

    • permutations: (1, 2, 3) (1, 3, 2) (2, 1, 3) (2, 3, 1), (3, 1, 2) (3, 2, 1)
       
    1. 1 precedes 3 is 3/6
        
    2. 3 precedes 1 is 3/6
        
    3. 3 precedes 1 and 3 precedes 2 is 2/6

     

  • 12 (4 pts)

    Suppose that E and F are events such that p(E) = 0.8 and p(F) = 0.6. Show that p(E U F) ³ 0.8 and p(E Ç F) ³ 0.4.
    
    Hint: for p(E Ç F) ³ 0.4, use Rule of Inclusion-Exclusion (Theorem 2)

    p(E U F) ³ p(E) = 0.8 because |E U F| ³ |E|

    p(E U F) £ 1
    p(E È F) = p(E) + p(F) - p(E Ç F)  £ 1 p(E È F) = 0.8 + 0.6  - p(E Ç F)  £ 1 1.4 - p(E Ç F)  £ 1 0.4 - p(E Ç F)  £ 0 0.4 £ p(E Ç F) p(E Ç F) ³ 0.4	Theorem 2 p397 For E1 and E2 events in sample space S, p( E1 È E2  ) = p( E1 ) + p( E2  )  - p( E1 Ç E2  )

     

  • 24 (2 pts)

    What is the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up tails?

    E is 4 Hs

    F is first flip a T   

    |S| = 32, the number of ways to flip 5 coins

    |F| = 16, the number of ways to flip remaining 4 coins after the first flip a T

    p(F) = 16/32

    E Ç F is THHHH

    |E Ç F| = 1

    p(E Ç F) = 1/32

    p(E | F) = p(E Ç F)/p(F)

                 = (1/32) / (16/32)

                 = (1/32)/(1/2)

                 = 1/16

• Section 6.3 -- pp. 424 - 425

   

  • 2 (2 pts)

    Suppose that E and F are events in a sample space and p(E) = 2/3 and p(F) = 3/4 and p(F|E) = 5/8. Find p(E|F).

    p(E | F) = p(E Ç F)/p(F)

    p(E Ç F) = p(E)p(F|E) = (2/3) (5/8) = 5/12

     p(E | F) = (5/12)/(3/4) = 20/36 = 5/9

  • 6 (4 pts)

    A test for steroids is given to soccer players.

    98% of the players using steroids test positive

    12% of the players not using steroids test positive.

    Suppose that 5% of soccer players use steroids.

    What is the probability that a soccer players who tests positive uses steroids?

     

    S is steroids are used by a random player, 5%.

    P is positive test for steroids by a random player.

    p(S) = 5% of players do take steroids

    p(S) = 95% of players do not take steroids

    p(P|S) = Given they use steroids, S, 98% test positive, P

    p(P|S) = Given they do not use steroids, S, 12% test positive, P

    p(S | P) = Given they test positive, P, the probability a player uses steroids, S.

    p(S | P) =

    p(P|S) p(S)              p(P|S) p(S) + p(P|S) p(S)

    p(S | P) =

    .98 * .05        =  0.301   .98*.05 +  .12*.95

    Given they test positive, P, 30% take steroids, S

     

• Section 6.4 -- pp. 439 - 440
  • 2 (1 pt)

    What is the expected number of heads that come up when a fair coin is flipped 10 times?

    Theorem 2: n * p

    n = 10

    p = 1/2

    n * p = 5

  • 4 (1 pt)

    A coin is biased so that the probability a head comes up when flipped is 0.6. What is the expected number of heads that come up when flipped 10 times?
     

    n * p = 10 * .6 = 6

  • 6 (2 pts)

    What is the expected value when a $1 lottery ticket is bought in which the purchaser wins $10,000,000 if the ticket contains the six winning numbers chosen from the set {1,2,...,50} and the purchaser wins nothing otherwise?

    probability of winning is 1/C(50, 6)

    Where losing costs $0.00

    E(X) = $10⁷ * 1/C(50,6) + $0.00 = $10⁷/C(50,6) = $0.63

    Where losing costs $1.00

     E(X) = $10⁷ * 1/C(50,6) -  $1 * ((C(50,6) – 1)/C(50,6)

            = 1/15890700 * $9,999,999 - $1 * 15890699/15890700

            = .6292988 - .999999 = -$0.37

  • (2 pts)    What is the variance and standard deviation of the number of times a heads comes up when a fair coin is flipped 4 times?

    |S| = 2⁴ = 16

    X random variable that assigns the number of heads flipped.

    h X(h) p(h) p(h)*X(h) HHHH 4 1/16 4/16 HHHT 3 1/16 3/16 HHTH 3 1/16 3/16 HTHH 3 1/16 3/16 THHH 3 1/16 3/16 HHTT 2 1/16 2/16 HTTH 2 1/16 2/16 TTHH 2 1/16 2/16

    h X(h) p(h) p(h)*X(h) HTHT 2 1/16 2/16 THTH 2 1/16 2/16 THHT 2 1/16 2/16 HTTT 1 1/16 1/16 THTT 1 1/16 1/16 TTHT 1 1/16 1/16 TTTH 1 1/16 1/16 TTTT 0 1/16 0/16

    E(X) = 1/16(1*4 + 4*3 + 6*2 + 4*1 + 1*0) = 1/16(32) = 2

    E(X²) = 1/16(4² + 4*3² + 6*2² + 4*1² + 0) = 1/16(16+36+24+4) = 1/16(80) = 5

    V(X) = E(X²) - E(X)² = 5 - 2² = 5 - 4 = 1

    s(X) = 1^1/2 = 1