# Why is the answer of the problem 0.57, rather than 1.57? According to me, the result should be $$1.57$$.

$$P(X \ge 1.8)$$
$$= F(4) - F(1.8) + 1$$ (as we have $$X>4 = 1$$)
$$= 1.57$$

Why is the answer of the problem $$0.57$$, rather than $$1.57$$?

• You expect the probability to be more than 1? More than certain? – badjohn Feb 28 at 13:11
• Why do you add one? I don't understand that. – Eff Feb 28 at 13:20
• @Eff, good question. $X \ge 1.8$ means the value of $X$ exceeds $4$. – user366312 Feb 28 at 13:22
• @badjohn, $X \ge 1.8$ means the value of $X$ exceeds $4$. So, why not include $1$? – user366312 Feb 28 at 13:24
• I considered an answer but others have done so now. – badjohn Feb 28 at 13:26

Yes, the answer should be $$0.57$$. First and foremost, the answer cannot be more than $$1$$. Thats a violation of conservation of probability. So the additional $$+1$$ that you mention in your answer is wrong and not needed, since $$F(X>4)=1$$ but $$P(X>4)=0$$. So you need not add the $$1$$. To prove this statement of mine, note that $$P(X>1.8)=F(X=4)-F(X=1.8)$$ $$=F(X=\infty)-F(X=1.8)$$

It means that the cumulative probability distribution reaches 1, i.e. the sum of all the probabilities reach 1 on crossing X=4. So the cumulative probability function saturates at X=4 with a value of 1. There is no further contribution to F from any point X>4. This implies that P=0 for X>4. And if P=0 in that region, then there is no need for you to add 1 for X>4.

Hope this helps.

• "... , since $F(X>4)=1$ but $P(X>4)=0$." --- what do you mean? Kindly rephrase. – user366312 Feb 28 at 13:21
• It means that the cumulative probability distribution reaches 1, i.e. the sum of all the probabilities reach 1 on crossing X=4. So the cumulative probability function saturates at X=4 with a value of 1. There is no further contribution to F from any point X>4. This implies that P=0 for X>4. And if P=0 in that region, then there is no need for you to add 1 for X>4. – SchrodingersCat Feb 28 at 13:25
• Kindly, add this comment to the main answer so that I can accept this. – user366312 Feb 28 at 13:32
• @user366312 Done. – SchrodingersCat Feb 28 at 13:41

The answer is simply \begin{align} P(X\geq 1.8) &= 1-P(X<1.8)\\ &= 1- F(1.8)\\ &= 1-\frac{1}{32}\left(6\cdot1.8^2-1.8^3\right)\\ &\approx 0.57 \end{align}

• $X \ge 1.8$ means the value of $X$ exceeds $4$. – user366312 Feb 28 at 13:24
• @user366312 I must admit that I'm not sure I understand what you're getting at.The value 1.8 does not exceed 4, it's less than 4. But even if it did, I don't understand why it's relevant. – Eff Feb 28 at 13:26
• I guess that he is thinking that $X \ge 1.8$ means that $X$ might exceed $4$. Of course, for this particular distribution it won't. – badjohn Feb 28 at 13:29
• @badjohn I suppose. But even if could exceed 4, I don't understand how it is relevant. I don't think I understand OP's reasoning. – Eff Feb 28 at 13:32
• @Eff I just wondering that it might be the source of his confusion not that I thought that he was right. Clearly he isn't since he calculated a probability of more than $1$. – badjohn Feb 28 at 13:35