The McTofu restaurant is running a contest where customers can match a symbol on

Question

The McTofu restaurant is running a contest where customers can match a symbol on the inside the wrapper "Big Tofu" sandwich to win either food or money prizes. The restaurant claims the probability of winning a food prize is 0.7, the probability of winning a cash prize from $1 to $5 is 0.2. Is this probability assignment legitimate?________ explain. The data shown here were collected via a phone survey in 2010 of 625 cell phones users of age 12 to 17 by the Pew Research Foundation. If we select a person at random from this sample, a. what is the probability that the person responded that s/he texted more than 100 times per day? b. what is the probability that the person responded that s/he texted between 21 and 100 times on a typical day c. what is the probability that the person did not respond that she never texts?

Explanation / Answer

Question 1:

The probability assignment is said to be legitimate if the assignment conforms the laws or the rules of the probability distribution.

Suppose, A, B and C are the events of the trial, then, probability assignment to these events is legitimate if and only if,

P(A), P(B) and P(C)should be independent of each other and strictly between [0,1].

And P(A) + P(B) + P(C) = 1

Now, in our case the trial is winning money or food prizes in a contest.

Here, we have three events,

Event (A) = Winning food prize

Event (B) = Winning cash prize from $1 to $5 and

Event (C) = Winning higher cash prize

Thus, as per given information,

P(A) = 0.7, P(B) = 0.2 and P(C) = 0.2.

Now if we see above rules of the probability distribution,

>P(A), P(B) and P(C)should be independent of each other

is satisfied

> and strictly between [0,1]

is also satisfied.

But,

> And P(A) + P(B) + P(C) = 1

is not satisfied as we have

And P(A) + P(B) + P(C) = 0.7+.02+.02 =1.1 which is greater than 1.

And hence this probability assignment is not legitimate as it is not conforming the rule of the probability distribution.

Question 2:

Suppose A be the event of the trial with total N observations.

Let n(A) be the number of observations associated with event A, then we define the probability of happening of event A as,

P(A) = n(A) / N .......(Number of observations associated with event divided by total number of observations)

Now for our example N = 625 (Total number of teen cell phone users)

Also we have the distribution as below,

.Now, note that the % given in the above distributions are actually nothing but the probabilities of the occurring a particular event.

Like, P( Sending or receiving 0 text messages on a particular day) = 0.02 (2% = 2/100 = 0.02)

similarly, P(Sending or receiving 11-20 text messages on a particular day) = 0.11 (11% = 11/100 = 0.11)

(a) Here we need to find the probability that the person responded that s/he texted more than 100 times per day.

If we look at the table, we see "?" there. So how to go about it?

Yes, here we need to use the rule of probability distribution. The addition of the probabilities should be equal to 1.

Thus here we have 6 events with their probabilities as,

P(0) = 0.02

P(1-10) = 0.22

P(11-20) = 0.11

P(21-50) = 0.18

P(51-100) = 0.18

P(100+) = ?

Now, we know that addition of this all probabilities should be 1.

Lets add it, 0.02 + 0.22+ 0.11+ 0.18+ 0.18 + ? = 1

which is, 0.71 + ? = 1

Hence, ? = 1-0.71 = 0.29

Thus, probability that the person responded that s/he texted more than 100 times per day is 0.29.

(b) The probability that the person responded that s/he texted between 21 and 100 times on a particular day will be addition of probabilities for (21-50) and (51-100).

The person who texted between 21 and 100 times must be from class 21-50 OR 51-100.

Thus, probability that the person responded that s/he texted between 21 and 100 times on a particular day =

P(21-50) + P(51-100) = 0.18 +0.18 = 0.36.

(c) Now, to find the probability that the person did not respond that she never texts, we first need to find the probability that the person responds that she never texts that is she texts 0 times.

Thus, P( 0 ) = 0.02.

Now , note that, If A is the event of happening something with probability P(A),

then the probability of not happening the event A is P(A') = 1 - P(A) .....(Read A' as A not)

Thus, probability that person texts 0 times is 0.02, thus probability that person do not texts 0 times is

P(0') = 1 - P(0)

= 1-0.02

= 0.98

Thus, probability that the person did not respond that she never texts is 0.98.

Number of text message sent or received on a typical day % of teen cell phone users 0 2% 1-10 22% 11-20 11% 21-50 18% 51-100 18% 101+ ?

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.