Ever had a pistachio nut Some of them are not cracked open a
Solution
Let X be the random variable representing the number of pistachios in th bag that cannot be opened by hand.
If two bags were selected at one after the other what is the probability of finding that both bags had no pistachios in the bag was too difficult to open by hand.
That is here we have to find P(bag1 has no pistachios in the bag was too difficult to open by hand. and bag2 has no pistachios in the bag was too difficult to open by hand.)
= P(bag1 no pistachios in the bag was too difficult to open by hand.) * P(bag2 has no pistachios in the bag was too difficult to open by hand.)
= P(X=0)*P(X=0 )= 0.5 * 0.5 = 0.25
If two bags were selected and the first bag was opened and had three pistachios in the bag that was too difficult to open what is the probability that the second bag has exactly three pistichios too difficult to open by hand.
P(X = 3) = 0.05
Suppose two bags are selected at random. What is the probability that both bags have four or more pistachios too difficult to open by hand?
P(bag1 have four or more pistachios too difficult to open by hand and bag2 have four or more pistachios too difficult to open by hand)
P(X >=4) * P(X >=4)
P(X = 4 or 5) * P(X = 4 or 5)
P(X=4)+P(X=5) * P(X=4)+P(X=5)
= 0.075+0.025 * 0.075+0.025
= 0.1*0.1 = 0.01
In the next part, we have to find mean and standard deviation the probability distribution.
mean = x*p(x)
variance = x2 *p(x) - mean2
sd = sqrt (variance)
The table for calculating mean and sd is, as follows.
mean = 1.03
variance = 2.9 - 1.032 = 1.839
sd = sqrt(1.839) = 1.36
| x | p | xp | x^2p |
| 0 | 0.5 | 0 | 0 |
| 1 | 0.25 | 0.25 | 0.25 |
| 2 | 0.1 | 0.2 | 0.4 |
| 3 | 0.05 | 0.15 | 0.45 |
| 4 | 0.075 | 0.3 | 1.2 |
| 5 | 0.03 | 0.125 | 0.625 |
| 1.03 | 2.9 |

