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givi [52]
3 years ago
13

A tennis player wins a match 55% of the time when she serves first and 47% of the time when her opponent serves first. The playe

r who serves first is determined by a coin toss before the match. What is the probability that the player wins a given match?
Mathematics
1 answer:
Ray Of Light [21]3 years ago
8 0

ok so she has a 50% chance of serving first so if she serves first she has a 55% percent chance of winning multiply .55 and .50 to get .275 now she also has a 50% chance of not serving first so multiply .47 and .50 to get .235 now that you have the two probabilities of her winning you add them together so .275+.235 = .51 so she has a <u>51%</u> chance of winning the game
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Thank you if you solve this I usually have a calculator....
prisoha [69]

nope its not a function

4 0
3 years ago
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How do you graph the function?
Bogdan [553]
Use you brain. Please use it.
4 0
2 years ago
Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive
PtichkaEL [24]
Looks like the PMF is supposed to be

\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65

Next,

\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25

\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5

\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2
\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5
\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}

If Y=X^2+1, then X^2=Y-1\implies X=\sqrt{Y-1}, where we take the positive root because we know X can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that Y can take on the values 1^2+1=2, 2^2+1=5, and 5^2+1=26. At these values of Y, we would have the same probability as we did for the respective value of X. That is,

\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}

Part (5) is incomplete, so I'll stop here.
8 0
3 years ago
What is 2 and 5/6 divided by 6 and 4/5
9966 [12]
Firstly I would revert these back to improper fractions:

So 17/6 ÷ 24/5

Then I would use keep, change, flip

So 17/6 * 5/24

Finally simplify

85/144

Final Answer: 85/144
3 0
3 years ago
Helppppppp??????!!!!!!!!!!
zzz [600]
I think the answer would be C. (btw, you should charge your phone/tablet, lol)
7 0
3 years ago
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