CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$(1+3p)/3,(1-p)/4$$ and $$(1-2p)/2$$ are the probabilities of three mutually exclusive events, then the set of all values of p is.............


Solution

Let,
$$P(A)=\dfrac{1+3p}{3},P(B)=\dfrac{1-p}{4},P(C)=\dfrac{1-2p}{2}$$
A,B and C are three mutually exclusive events.
$$\therefore P(A)+P(B)+P(C)\leq 1$$
$$\Rightarrow \dfrac{1+3p}{3}+\dfrac{1-p}{4}+\dfrac{1-2p}{2}\leq 1 \leq 1$$
$$\Rightarrow 4+12p+3-3p+6-12p \leq 12$$
$$\Rightarrow p \leq 1/3$$................(1)
Also,
$$0 \leq P(A) \leq 1 \Rightarrow 0 \leq \dfrac{1+3p}{3}\leq 1$$
$$0 \leq 1+3p \leq 3$$
$$\Rightarrow -\dfrac{1}{3}\leq p \leq \dfrac{2}{3}$$............(2)
$$0\leq P(B)\leq 1\Rightarrow 0 \leq \dfrac{1-p}{4}\leq 1$$
$$\Rightarrow 0 \leq 1-p\leq 4$$
$$\Rightarrow -3 \leq p \leq 1$$.................(3)
$$0 \leq P(C)\leq 1\Rightarrow 0 \leq \dfrac{1-2p}{2}\leq 1$$
$$\Rightarrow -\dfrac{1}{2}\leq p \leq \dfrac{1}{2}$$.................(4)
Combining eq.(1),(2),(3) and (4) ,we get
$$\dfrac{1}{3}\leq p \leq \dfrac{1}{2}$$


Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image