Chapter 5.2, Problem 33E

To determine

(a)

To find: The probability distribution of the data

Answer to Problem 33E

The probability distribution is

  $\text{X }~\text{ Bin }\left(\text{6 },\frac{1}{6}\right)$

P (X = 2) = $\times {{\displaystyle \left(\frac{1}{6}\right)}}^2\times {{\displaystyle \left(1-\frac{1}{6}\right)}}^{6-2}$

Explanation of Solution

Given information:

Number of grandchildren = 6

Number of children to be picked for the tickets = 2

Let X be the number of grandchildren to be selected.

As each grandchild has an equal of selection, so the probability of each grandchild to be selected is $\frac{1}{6}$ .

There are in total 6 grandchildren are independent of each other.

So, $\text{X }~\text{ Bin }\left(\text{6 },\frac{1}{6}\right)$

  $\text{P (X = x ) = }\times {{\displaystyle \left(\frac{1}{6}\right)}}^x\times {{\displaystyle \left(1-\frac{1}{6}\right)}}^{6-x}\text{ for x = 1, 2, 3, 4, 5, 6}$

To determine

(b)

To compute: Probability that at least one of the Piper or Sedona get the tickets.

Answer to Problem 33E

Probability that at least one of the Piper or Sedona get the tickets is 0.6.

Explanation of Solution

Given information:

The information as mentioned in the above sub - part.

Formula used:

P (at least one of the girls get the tickets) = P (one of the girls (Piper and Sedona)) + P (both of the girls (Piper and Sedona))

Calculation:

P (one of the girls (Piper and Sedona)) = $\frac{}{6{{\displaystyle C}}_2}=0.5333$

P (both of the girls (Piper and Sedona)) = $\frac{}{6{{\displaystyle C}}_2}=0.0667$

P (at least one of the girls get the tickets) =0.5333 + 0.0667 = 0.6