Proby quest.s giving me problems........DOES NE ONE LIKE PROBABILITY NE WAY? (1 Viewer)

[lillaila]

*sigh*.........i hate these questions......i hope u can help, my trials are in a few days

1. In a train compartment, there r 8 seats, with 4 facing the front and 4 facing backwards.
a) If 2 of the pple do not like sitting backwards, in how many ways can the 5 pple b arranged?................ans: 1440
b) Find the proby that 2 particular pple will sit opposite each other is seating is arranged at random................ans: 1/7

2. A game of poker uses a deck of 52 cards with 4 suits (hearts, diamonds, spades and clubs). Each suit has 13 cards, consisting of an ace, cards numbered from 2 to 10, a jack, queen and king. If a person is dealt 5 cards find the proby of getting :
a) 4 aces............ans: 1/54145
b) a flush (all cards the same suit)......ans: 33/16660

AND THIS STUPID Q.!.....i doubt many or any will get this

3. I throw a coin "k" times. Find an expression to describe the proby of throwing:
a) at least one tail................ans: 1-(1/2*k)
b) (k-3) heads..........ans: (*kC_k-3)(1/2*k)
c) 9 tails.........ans: (*kC_9)(1/2*k)


Thank You

 

[Constip8edSkunk]

q1
2 people can sit in 4 seats and 3 ppl can sit in any of the remaining seats:
4*3*6*5*4=1440

P(2 ppl opp.)
= arrangements of 2 particular ppl sitting opp. / total arrangements
=8*6*5*4/8*7*6*5*4(set 2 seats aside and allocate the remaining 3, then multiply by 8 as there are 8 different possible arrangements for the 2ppl)
=1/7

q2
P(4A) = 48/52C5 = 1/54145

P(flush) = 4*[1/4*12/51*11/50*10/49*9/58] = 33/16660

q3
P(at least 1T) = 1 - P(0 T)
=1 - 0.5^k

P(k-3 H) = kC(k-3)*0.5^k*0.5^[k-(k-3)] _{(using binomial prob.)}
=kC(k-3)*0.5^[k-3+3)]
=kC(k-3)*0.5^k

P(9T) = kC9*0.5^9*0.5^k-9
=kC9*0.5^[9+k-9]
=kC9*0.5^k

 

[Affinity]

alternative 1b)

let the 2 trouble makers be A and B
let A sit down.
there will be 7 seats remaining
and only if B is in that seat will the conditions be satisfied.
therefore 1/7