Expected Value: Use Indicator variables and random variables
$begingroup$
Question:
Every time a customer orders a drink, the waiter serves the wrong drink with probability $frac{1}{12}$, independently of other orders.
You order $7$ ciders, one cider at a time. Let $(D_1,D_2,...,D_7)$ be the sequence of drinks that the waiter serves. Define the following random variable $X$:
$X$ = the number of indices i such that $D_i$ is a cider and $D_{i+1}$ is not a cider.
What is the expected value $E(X)$ of $X$
Answer: 0.45833333
Attempt:
I start off labelling my indicator variable:
$$
X = left{begin{array}{rc} 1,&text{the number of indices i such that $D_i$ is a cider and $D_{i+1}$ is not a cider}{} \ 0,&text{any other cases}{}end{array}right.
$$
I need to find $P(X=1)$ but I'm not sure how to go about it. Will I be served 4 ciders and 3 non-cider drinks according to the condition? How do I incorporate the probability of the waiter getting the drink wrong?
I am struggling to proceed with these questions after defining the indicator variables. Any step by step guide would be appreciated.
probability-theory discrete-mathematics random-variables expected-value
asked Dec 1 '18 at 6:10
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
Question:
Every time a customer orders a drink, the waiter serves the wrong drink with probability $frac{1}{12}$, independently of other orders.
You order $7$ ciders, one cider at a time. Let $(D_1,D_2,...,D_7)$ be the sequence of drinks that the waiter serves. Define the following random variable $X$:
$X$ = the number of indices i such that $D_i$ is a cider and $D_{i+1}$ is not a cider.
What is the expected value $E(X)$ of $X$
Answer: 0.45833333
Attempt:
I start off labelling my indicator variable:
$$
X = left{begin{array}{rc} 1,&text{the number of indices i such that $D_i$ is a cider and $D_{i+1}$ is not a cider}{} \ 0,&text{any other cases}{}end{array}right.
$$
I need to find $P(X=1)$ but I'm not sure how to go about it. Will I be served 4 ciders and 3 non-cider drinks according to the condition? How do I incorporate the probability of the waiter getting the drink wrong?
I am struggling to proceed with these questions after defining the indicator variables. Any step by step guide would be appreciated.
probability-theory discrete-mathematics random-variables expected-value
asked Dec 1 '18 at 6:10
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
Question:
Every time a customer orders a drink, the waiter serves the wrong drink with probability $frac{1}{12}$, independently of other orders.
You order $7$ ciders, one cider at a time. Let $(D_1,D_2,...,D_7)$ be the sequence of drinks that the waiter serves. Define the following random variable $X$:
$X$ = the number of indices i such that $D_i$ is a cider and $D_{i+1}$ is not a cider.
What is the expected value $E(X)$ of $X$
Answer: 0.45833333
Attempt:
I start off labelling my indicator variable:
$$
X = left{begin{array}{rc} 1,&text{the number of indices i such that $D_i$ is a cider and $D_{i+1}$ is not a cider}{} \ 0,&text{any other cases}{}end{array}right.
$$
I need to find $P(X=1)$ but I'm not sure how to go about it. Will I be served 4 ciders and 3 non-cider drinks according to the condition? How do I incorporate the probability of the waiter getting the drink wrong?
I am struggling to proceed with these questions after defining the indicator variables. Any step by step guide would be appreciated.
probability-theory discrete-mathematics random-variables expected-value
asked Dec 1 '18 at 6:10
TobyToby
1577
$endgroup$
Question:
Every time a customer orders a drink, the waiter serves the wrong drink with probability $frac{1}{12}$, independently of other orders.
You order $7$ ciders, one cider at a time. Let $(D_1,D_2,...,D_7)$ be the sequence of drinks that the waiter serves. Define the following random variable $X$:
$X$ = the number of indices i such that $D_i$ is a cider and $D_{i+1}$ is not a cider.
What is the expected value $E(X)$ of $X$
Answer: 0.45833333
Attempt:
I start off labelling my indicator variable:
$$
X = left{begin{array}{rc} 1,&text{the number of indices i such that $D_i$ is a cider and $D_{i+1}$ is not a cider}{} \ 0,&text{any other cases}{}end{array}right.
$$
I need to find $P(X=1)$ but I'm not sure how to go about it. Will I be served 4 ciders and 3 non-cider drinks according to the condition? How do I incorporate the probability of the waiter getting the drink wrong?
I am struggling to proceed with these questions after defining the indicator variables. Any step by step guide would be appreciated.
probability-theory discrete-mathematics random-variables expected-value
probability-theory discrete-mathematics random-variables expected-value
asked Dec 1 '18 at 6:10
TobyToby
1577
asked Dec 1 '18 at 6:10
TobyToby
1577
asked Dec 1 '18 at 6:10
TobyToby
1577
asked Dec 1 '18 at 6:10
TobyToby
1577
asked Dec 1 '18 at 6:10
TobyToby
1577
1577
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let $X_i$ be indicator variable for $i$-th event, so $X=X_1+...+X_7$. Then , for $ileq 6$ we have $$E(X_i)=P(X_i=1) = {11over 12}cdot {1over 12} $$ and $$E(X_7)= P(X_7 = 1)= {11over 12}cdot 0 = 0$$
so $$E(X) = E(X_1)+E(X_2)+...+E(X_7) =6cdot {1over 12}cdot {11over 12} = {11over 24} $$
answered Dec 1 '18 at 6:22
greedoidgreedoid
42.2k1152105
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $X_i$ be indicator variable for $i$-th event, so $X=X_1+...+X_7$. Then , for $ileq 6$ we have $$E(X_i)=P(X_i=1) = {11over 12}cdot {1over 12} $$ and $$E(X_7)= P(X_7 = 1)= {11over 12}cdot 0 = 0$$
so $$E(X) = E(X_1)+E(X_2)+...+E(X_7) =6cdot {1over 12}cdot {11over 12} = {11over 24} $$
answered Dec 1 '18 at 6:22
greedoidgreedoid
42.2k1152105
$endgroup$
add a comment |
$begingroup$
Let $X_i$ be indicator variable for $i$-th event, so $X=X_1+...+X_7$. Then , for $ileq 6$ we have $$E(X_i)=P(X_i=1) = {11over 12}cdot {1over 12} $$ and $$E(X_7)= P(X_7 = 1)= {11over 12}cdot 0 = 0$$
so $$E(X) = E(X_1)+E(X_2)+...+E(X_7) =6cdot {1over 12}cdot {11over 12} = {11over 24} $$
answered Dec 1 '18 at 6:22
greedoidgreedoid
42.2k1152105
$endgroup$
add a comment |
$begingroup$
Let $X_i$ be indicator variable for $i$-th event, so $X=X_1+...+X_7$. Then , for $ileq 6$ we have $$E(X_i)=P(X_i=1) = {11over 12}cdot {1over 12} $$ and $$E(X_7)= P(X_7 = 1)= {11over 12}cdot 0 = 0$$
so $$E(X) = E(X_1)+E(X_2)+...+E(X_7) =6cdot {1over 12}cdot {11over 12} = {11over 24} $$
answered Dec 1 '18 at 6:22
greedoidgreedoid
42.2k1152105
$endgroup$
Let $X_i$ be indicator variable for $i$-th event, so $X=X_1+...+X_7$. Then , for $ileq 6$ we have $$E(X_i)=P(X_i=1) = {11over 12}cdot {1over 12} $$ and $$E(X_7)= P(X_7 = 1)= {11over 12}cdot 0 = 0$$
so $$E(X) = E(X_1)+E(X_2)+...+E(X_7) =6cdot {1over 12}cdot {11over 12} = {11over 24} $$
answered Dec 1 '18 at 6:22
greedoidgreedoid
42.2k1152105
edited Dec 1 '18 at 6:27
answered Dec 1 '18 at 6:22
greedoidgreedoid
42.2k1152105
answered Dec 1 '18 at 6:22
greedoidgreedoid
42.2k1152105
answered Dec 1 '18 at 6:22
greedoidgreedoid
42.2k1152105
42.2k1152105
add a comment |
add a comment |
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