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June 16th, 2016, 08:31 AM   #1
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Probability

Can someone help me and explain how to do this question?

Last edited by jamesbrown; June 16th, 2016 at 08:41 AM.
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June 16th, 2016, 08:50 AM   #2
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Can you show us any work you've done/any previous attempt?
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June 16th, 2016, 09:22 AM   #3
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Yeah i dont even know where to start so I don't have any working?
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June 16th, 2016, 12:28 PM   #4
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If, as you appear to be saying, you have never studied probability, where did you get this problem?

If the probability of something happening is 'p' then the probability of it happening twice is $\displaystyle p^2$. Here, since the probability of "heads, twice" is 0.36 so the probability of "heads" is 0.6. The probability of a single tails, then, is 1- 0.6= 0.4. The probability of "tails, twice" is $\displaystyle (0.4)^2= 0.16$.

For the second problem you are told that the probability of exactly one head when a coin is tossed twice is 8/25. If the probability of "heads" on one toss is p, then the probability of a tail is 1- p. The probability of one head when tossed twice is $\displaystyle 2p(1- p)= 2p- 2p^2= 8/25$. Solve $\displaystyle 2p^2- 2p+ 8/25= 0$ for p. Then calculate $\displaystyle 25p^2- 25p+ 4$.

If there are n red beads out of 10 then the probability the first drawn is red is $\displaystyle \frac{n}{10}$. If Meg draws a red bead and does not replace it, there are now 9 beads left, n-1 red. The probability of getting a red bead on the second draw is $\displaystyle \frac{n-1}{9}$. The probability of drawing two red beads is ]math]\frac{n(n-1)}{90}= \frac{1}{3}[/math]. Solve that for n and calculate $\displaystyle n^2- n- 30$.

Let P be the number of people in the club. We are told that 1/3 of the people in the club are men and that is "n". So n= (1/3)P and P= 3n.

If there are n men out of P people in the club then the probability one person chosen at random is a man is $\displaystyle \frac{n}{P}$. That leaves n-1 men out of P-1 people so the probability the second person chosen is also man is $\displaystyle \frac{n-1}{P-1}$. So the probability two people chosen are both men is $\displaystyle \frac{n(n-1)}{P(P-1)}= \frac{1}{10}$. But, as above, P= 3n so $\displaystyle \frac{n(n-1)}{3n(3n-1)}= \frac{1}{10}$. Solve that for n and then calculate P.

If these questions were for a course and you really "have no idea where to start" you need to talk to your teacher!

Last edited by Country Boy; June 16th, 2016 at 01:27 PM.
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June 16th, 2016, 01:24 PM   #5
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$\displaystyle \frac{n(n-1)}{90}= \frac{1}{3}$
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June 17th, 2016, 07:47 PM   #6
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Multiple n(n - 1). Then cross-multiply the fractions. You will get a quadratic equations that can be factored.
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