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A random variable X has a binomial distribution with parameters n = 4 and p. What must be the value of the parameter p if we know that P(X = 3)= P(X = 4)?

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can you help?
please help
\[P(X=3)=\left(\begin{matrix}4 \\ 3\end{matrix}\right)p ^{3}(1-p)=4p ^{3}(1-p)=4p ^{3}-4p ^{4}\] \[P(X=4)=\left(\begin{matrix}4 \\ 4\end{matrix}\right)p ^{4}(1-p)^{0}=p ^{4}\] \[4p ^{3}-4p ^{4}=p ^{4}\ ....................(1)\] Now solve equation (1) to find the value of p.

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Other answers:

i cant solve that euqation?
Do you want me to help you to solve the equation?
\[4p ^{3}-4p ^{4}=p ^{4}\] First add 4p^4 to both sides.
What did you get?
4p^3= 5p^4
Now divide both sides by 5p^3 and post the result.
i dont get it
\[\frac{4p ^{3}}{5p ^{3}}=\frac{5p ^{4}}{5p ^{3}}\] Now simplify.
Yes, p = 4/5.
but the answer is 3/4
If you test the value of p = 3/4 you will find that it is not correct. Testing the value of p = 4/5 shows that it is correct. Did you copy the original question correctly?
yes i did
anyway, i gotthe idea, can I ask you one more question, i just need an explanation, please
A variable X follows a normal distribution with mean 10 and standard deviation 5. Another variable Y follows a normal distribution with mean 25 and standard deviation 10. The maximum height of the density curve for X is_______ (i) the maximum height for the density curve for Y , and the area under the density curve for X is__________ (ii) the area under the density curve for Y . (A) (i) greater than, (ii) less than (B) (i) less than, (ii) greater than (C) (i) equal to, (ii) equal to (D) (i) greater than, (ii) equal to (E) (i) less than, (ii) less than
Here is the checking of the answer p = 4/5: \[P(X=3)=\left(\begin{matrix}4 \\ 3\end{matrix}\right)\times (\frac{4}{5})^{3}\times \frac{1}{5}=(\frac{4}{5})^{4}\] \[P(X=4)=\left(\begin{matrix}4 \\ 4\end{matrix}\right)\times (\frac{4}{5})^{4}\times (\frac{1}{5})^{0}=(\frac{4}{5})^{4}\]
could you explain that one please
@yashar806 Please post your second question again after closing this one. Sorry I cannot help any further. I must log out now.

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