(急!!!)F5 Probability

1.
A)
N : Nick draws a card with a prime number
n : Nick draws a card without a prime number
M : Margaret draws a card with a prime number
m : Margaret draws a card without a prime number

Among the 9 cards, there are 2 cards with prime numbers (23, 29).
P(N) = P(M) = 2/9
P(n) = P(m) = 1 - (2/9) = 7/9

P(Nick will win)
= P(N) + P(nmN) + P(nmnmN) + P(nmnmnmN) + ...... to ∞terms
= (2/9) + (7/9)²(2/9) + (7/9)⁴(2/9)+ (7/9)⁶(2/9)+ ...... to ∞ terms
= (2/9) + (49/81)(2/9) + (49/81)²(2/9) + (49/81)³(2/9)+ ...... to ∞ terms

This is the sum of a geometric sequence to ∞terms with first term a = 2/9 and common ratio r = 49/81.
Sum of a geometric sequence to ∞ terms = a / (1 - r)

Hence, P(Nick will win)
= (2/9) / [1 - (49/81)]
= (2/9) / (32/81)
= (2/9) x (81/32)
= 9/16

B)
Method 1 :
P(Margaret will win)
= 1 - P(Nick will win)
= 1 - (9/16)
= 7/16

Method 2 :
P(Margaret will win)
= P(nM) + P(nmnM) + P(nmnmnM) + P(nmnmnmnM) + ...... to ∞ terms
= (7/9)(2/9) + (7/9)³(2/9) + (7/9)⁵(2/9)+ (7/9)⁷(2/9)+ ...... to ∞ terms
= (14/81) + (49/81)(14/81) + (49/81)²(14/81) + (49/81)³(14/81)+ ...... to ∞ terms

This is the sum of a geometric sequence to ∞terms with first term a = 14/81 and common ratio r = 49/81.

Hence, P(Margaret will win)
= (14/81) / [1 - (49/81)]
= (14/81) / (32/81)
= (14/81) x (81/32)
= 7/16


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2.
P($20 in one draw)
= P(two $10-coins)
= 50C2 / 60C2
= (50!/48!2!) / (60!/58!2!)
= (50 x 49) / (60 x 59)
= 245/354

P($11 in one draw)
= P($10-coin + one $1 coin)
= 50C1 x 10C1 / 60C2
= (50!/49!1!) x (10!/9!1!) / (60!/58!2!)
= (50 x 10) / (30 x 59)
= 50/177

P($2 in one draw)
= P(two $2-coins)
= 10C2 / 60C2
= (10!/8!2!) / (60!/58!2!)
= (10 x 9) / (60 x 59)
= 9/354

The required probability
= P($20, $11) + P($11, $20) + P($20, $2) + P($2, $20) + P($11, $2) + P($2, $11)
= 2 x P($20,$11) + 2 x P($20, 2) + 2 x P($11, $2)
= 2 x (245/354) x (50/177) + 2 x (245/354) x (9/354) + 2 x (50/177) x (9/354)
= (24500/62658) + (2205/62658) + (900/62658)
= 27605/62658