MHB Answer:Finding Value of k & Probability of A Winning at Least 2 Matches

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In a series of 10 ping-pong matches, player A has a winning probability of 0.2. For part a), the probability of A winning exactly k matches does not match the given probability of 0.888 for any integer k, suggesting a possible error in the problem statement. For part b), the probability that A wins at least two matches can be calculated by subtracting the probabilities of A winning zero or one match from 1. The correct approach involves using the binomial probability formula to compute these probabilities accurately. The discussion highlights the need for careful calculation in probability problems.
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Two ping-pong players $A$ & $B$ play a series of 10 matches. The probability that $A$ wins any of the matches is $0.2$

a) If the probability that $A$ wins exactly $k$ matches in the series is $0.888$. What's the value of $k$?

b) What's the probability that $A$ wins at least two matches?

---------------

I ain't good with these stuff, so I hope you guys may help me please. :D
The part a) states that $A$ could win 2, 3, 5 or 8 matches and the probability is the same, is this assumption correct? How to find the value of $k$?

As for b), how can I compute the asked probability? I know that $A$ has probability of wining any of the patches with value $0.2$, but when it says "at least," I can understand it.

Thank you.
 
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Kudasai said:
Two ping-pong players $A$ & $B$ play a series of 10 matches. The probability that $A$ wins any of the matches is $0.2$

a) If the probability that $A$ wins exactly $k$ matches in the series is $0.888$. What's the value of $k$?

b) What's the probability that $A$ wins at least two matches?

---------------

I ain't good with these stuff, so I hope you guys may help me please. :D
The part a) states that $A$ could win 2, 3, 5 or 8 matches and the probability is the same, is this assumption correct? How to find the value of $k$?

As for b), how can I compute the asked probability? I know that $A$ has probability of wining any of the patches with value $0.2$, but when it says "at least," I can understand it.

Thank you.

The probability that A wins k games of the 10 games is... $\displaystyle P \{k=0 \} = \binom {10}{0}\ .2^{0}\ .8^{10} = .10737...$

$\displaystyle P \{k=1 \} = \binom {10}{1}\ .2\ .8^{9} = .2684...$

$\displaystyle P \{k=2 \} = \binom {10}{2}\ .2^{2}\ .8^{8} = .3019...$

$\displaystyle P \{k=3 \} = \binom {10}{3}\ .2^{3}\ .8^{7} = .201326...$

... and it is fully evident that P= .888 doesn't correspond to any value of k. The probability $P\{k \ge 2\}$ is... $\displaystyle P\{k \ge 2\} = 1 - P\{k = 0\} - P\{k= 1\}\ (1)$

Kind regards

$\chi$ $\sigma$
 
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I agree with chisigma that for part a) there is no solution, however, I suspect the given probability should be approximately 0.088. Then we may write:

$${10 \choose k}(0.2)^k(0.8)^{10-k}\approx0.088$$

Try successive values for $k$.

For part b) I would use the fact that:

$$P(\text{A wins less than two})+P(\text{A wins at least two})=1$$

We can arrange this as:

$$P(\text{A wins at least two})=1-P(\text{A wins zero OR A wins 1})$$

Can you proceed?
 
Okay so I have for b) that

$P(A)=1-P(A^c),$ so $P(A)=1-0.2=0.8.$ Is it correct? I think it's too easy to be correct.

Thanks.
 
Kudasai said:
Okay so I have for b) that

$P(A)=1-P(A^c),$ so $P(A)=1-0.2=0.8.$ Is it correct? I think it's too easy to be correct.

Thanks.

Yes, that's not correct. You want:

$$P(\text{A wins at least two})=1-P(\text{A wins zero OR A wins 1})$$

Now, we may write:

$$P(\text{A wins zero OR A wins 1})=P(\text{A wins zero})+P(\text{A wins one})$$

And apply the binomial probability formula to each event on the right.
 
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