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beads are taken at random oh and yes as uv guessed...non-calc.Also anuther similar one,A coin has been alteredso the probability of gtting a head has been reduced.When coin is tossed twice the probability of gttn one head,P(H,T)=P(T,H) is 0.32.what is the prob of gttn a head when the coin is tossed once? (trial&improvmnt nt taken) thnx

2007-05-26 12:03:33 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

Sorry it's taken so long-- I had to think back more years than I care to remember.

1. (P(R,R)=15/22). You didn't say whether the bead was replaced after the first draw. I first assumed it was. That meant P(rr) = x^2 / (x + 2)^2 = 15/22. That assumption did not produce an integer answer. You have to have an integer number of beads. So…

Assume the beads are not replaced after the first draw. There are x red beads in the pot for first draw and x-1 red beads for the 2nd draw. Also, there are x + 2 total beads for the first draw and x + 1 total beads for the second draw.

P(r1) is the probability of a red on the first draw= (x / x+2).
P(r2) is the probability of a red on the 2nd draw= (x-1 / x+1).
So P(rr) = P(r1) * P(r2) = [x (x - 1)] / [(x + 2)(x + 1)] = 15/22
Turn the crank and you get x = 10 or a fractional bead which is impossible.
Check: P(r) = 10/12 = 5/6 and P(rr) = (5/6)(9/11) = 15/22
Whoopee!

2. Contrary to your statement, the altered coin problem is not like the first bead problem. We don't have to worry about different numbers of outcomes on each flip.

We know P(1 head) = P(ht) or P(th) = 2P(ht) = .32. so
P(ht) = .16.

Also, P(t) + P(h) = 1 and P(t) = 1 - P(h)
P(ht) = P(h)* P(t) = P(h)*[1- P(h)] = P(h) - P(h)^2 = .16

Solving for P(h) gives 0.8 or 0.2. Since without alteration, P(h) is 0.5 AND since the alteration reduced the probability of getting a head, P(h) = 0.2

Ta - Dah!

Thanks for the mental exercise

2007-05-26 15:05:39 · answer #1 · answered by davec996 4 · 0 0

As stated above, neither problem has a valid solution. Please restate the problems exactly as they were given to you and resubmit.

2007-05-26 20:55:09 · answer #2 · answered by Northstar 7 · 0 0

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