Calculating Probability and Variance for Randomly Opened Tins

[jinx007]

I am having problem to calculate some probability...please help me and explain to me how you get the answer

In order to use the labels in a competition a housewife removes the labels from 3 tins of tomato soup and from 4 tins of peaches, but forget to mark the tins which, without their labels are all identical. The housewife opens successive tins, chosen at random, looking for a tin of peaches. The random variable X denotes the number of tins opened up to and including the first tin of peaches.

Tabulate the probability distribution of X and show that var(X) = 16/25


Please help me i just don't know where to start..

 

[praharmitra]

Question says we must draw up a probability distribution table of X = the number of tins one must open to get a peach. So let's start at that...

Say X = 1. So, we need to find the probability that in the very first try we open a peach tin. Can you find this?

When X = 2, we need the probability that the first tin we opened was NOT peach, but the second tin was.

For X = 3, find probability that the first two tins opened were tomatos, and the third one was peach...and so on

Similarly, you find all the probabilities, till X = 4 (Why 4??)

Then draw a table. That is your answer. Then find var(X) with the formula.

 

[jinx007]

Question says we must draw up a probability distribution table of X = the number of tins one must open to get a peach. So let's start at that...

Say X = 1. So, we need to find the probability that in the very first try we open a peach tin. Can you find this?

When X = 2, we need the probability that the first tin we opened was NOT peach, but the second tin was.

For X = 3, find probability that the first two tins opened were tomatos, and the third one was peach...and so on

Similarly, you find all the probabilities, till X = 4 (Why 4??)

Then draw a table. That is your answer. Then find var(X) with the formula.

Hmmmm i understand it partially. But i think that if my table is correct that is the total probability = 1 so i can manage the variance

for the probability of of getting 1 peach tin

P(x=1) = (4/7)

Getting it on the 2nd trial
P(x=2) = ( 3/7 x 4/6) = (2/7)

Getting it on the 3th trial
P(x=3) = ( 3/7 x 2/6 x 4/5) = (4/35)

Getting it on the fourth trial
P(x=4) = ( 3/7 x 2/6 x 1/5 x 4/4) = (1/35)

SO adding all the probability give me 1 (So my answer must be correct..hehehe

 

[praharmitra]

correct