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

[Kudasai]

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?

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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.

 

[chisigma]

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

$\chi$ $\sigma$

 

[MarkFL]

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?

 

[Kudasai]

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.

 

[MarkFL]

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.