A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P[G], where G is the event that a number greater than 3 occurs on a single roll of the die.
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P[G], where G is the event that a number greater than 3 occurs on a single roll of the die.
Solution:
The
probability of odd numbers
$\begin{array}{l}
=2 \times \text { [Probability of even number] } \\
\Rightarrow P[\text { Odd }]=2 \times P \text { [Even] } \\
\text { Now, } P[\text { Odd }]+P[\text { Even }]=1 \\
\Rightarrow 2 P \text { [Even }]+P[\text { Even }]=1 \\
\Rightarrow 3 P \text { [Even] }=1 \\
P[\text { Even }]=1 / 3
\end{array}$
Therefore,
$\mathrm{P}[\mathrm{O} \mathrm{dd}]=1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}$
The total no. occurs on a
single roll of die $=6$
And the no. greater than $3=4,5$ or 6
Therefore, $P[G]=P$ [no. greater than 3$]$
$=P[$ no. is 4,5 or 6$]$
So here, 4 and 6 are the even numbers and 5 is odd
$\begin{array}{l}
\therefore \mathrm{P}[\mathrm{G}]=2 \times \mathrm{P}[\text { Even }] \times \mathrm{P}[\text { Odd }] \\
=2 \times 1 / 3 \times 2 / 3 \\
=4 / 9
\end{array}$
As a result, the required probability is $4 / 9$
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A die is loaded in such a way that an even number is twice as likely to occur as an odd number: If E is the event that & number less than 4 occurs on a single toss of the die, find P[E]. Select one; 5/8 6/9 3/6 4/7
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Video Transcript
in this question, we are given that a dye is loaded in such a way that an even number is twice as likely to occur as an odd number. Now we know that a die is going to have these Numbers. 1, 2, 3, 4, 5 up to six. And the probability of having an even number, we are going to call it X. And the probability of having an odd number, we're going to call it. Why? Now? We are given that the probability of an even number X is twice the probability of odd number. Why? So we have a question one. And secondly we know that for this day we have our even numbers they said to four and six in our own numbers As he said one, three and five. No, We know that the probability of these numbers should add up to one if we throw a die once. So the probability of having probability of having any number that is That gives us one is equal to the probability Of it being a one Plus the probability of it being a two plus probability of it being a three plus. The probability of it being a four plus, the probability of it being a five plus lastly the probability of it being a six. Now again here we know that the probability of an odd number is we have called it. Why? So here we have Why? Plus X. Plus why? Plus this will be Y. X. Y. X. Plus why? Plus X. So and it gives us a one. So we have another equation with three X plus three Y is equal to one. So we'll call this equation to. Now solving these two equations three X plus three Y is equal to one and X is equal to two Y. Now replacing X by two I we have three by two. Y plus three Y is equal to one. Here we have six Y plus three Y Is equal to one. nine Y is equal to one and why is equal to 1/9. Therefore X which is two, Y is equal to 2/9. Now our question the event, the probability of having a number that is less than four, That is less than four is equal to the probability of it being a one plus The probability of it being a two Plus the probability of it being a three. And here we now know that the probabilities the probability of it being an odd number is 1/9 plus 2/9 plus 1/9. And therefore we get, our final answer is 4/9 which is not there on our given answers. And the question is um the options that are given, there's no right answer the correct answer is for overnight. Please contact your instructor. Thank you
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A die is loaded in such a way that an even number is twice as likely to occur as an odd number.
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A die is loaded in such a way that an even number is twice as likely to occur as an odd number. If E is the event that a number less than 4 occurs on a single toss of the die, find P[E].
Guest Feb 29, 2020
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\[P[E]=2P[O]\\ P[E]+P[O]=1\\ P[E]=\dfrac 2 3,~P[O]=\dfrac 1 3\\ P[1]=P[3]=P[5]=\dfrac 1 3 \cdot \dfrac 1 3 =\dfrac 1 9\\ P[2]=P[4]=P[6]=\dfrac 1 3 \cdot \dfrac 2 3 = \dfrac 2 9\\ P[roll < 4]=P[1]+P[2]+P[3] = \dfrac 1 9 + \dfrac 2 9 + \dfrac 1 9 = \dfrac 4 9\]
.Rom Mar 1, 2020
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Trial | Outcomes | Examples of Events |
Rolling a die | There are 6 possible outcomes: {1, 2, 3, 4, 5, 6} | Rolling an even number: {2, 4, 6} Rolling a 3: {3} Rolling a 1 or a 3: {1, 3} Rolling a 1 and a 3: { } [Only one number can be rolled, so this outcome is impossible. The event has no outcomes in it.] |