Hello I would like someone to help me check if the following answer is correct:

Question

Hello I would like someone to help me check if the following answer is correct:

You have a bag that contains 3 apples, 2 billiard balls, 7 avocados, and 3 baseballs. If you were to remove 2 items from the bag at random (so you do not know what kind of item is being removed) what are the chances that:

The items are both avocados (if the first item is replaced after its type is noted)

The items are both baseballs (if the first item is not replaced after its type is noted)

One item is a billiard ball and the other is an avocado (in any order)

Both items are food

n = total number of items = 3 +2 +7 + 3 = 15

a) The items are both avocados

The probability of first choosen item is avacodos = 7/15

Since we choose one item is randomly one by one with replacemnet . i.e .we select the item and type is noted , then the item replaced in the before the next selection. hence the the number of items in the bag are same at every draw.

The probability of second choosen item is avacodos = 7/15

Hence

P( The items are both avacados ) = 7/15 * 7/15 = 49/225 = 0.2177

P( The items are both avacados ) = 0.2177

b) The items are both baseballs

Probability first choosen item is baseball = 3/15

Since the replication is not allowed. i.e. the number of items in abag are reduced by one at every draw.

Probability Second choosen item is baseball = 2/14

P( The items both are baseball) = 3/15 *2/14 = 6/210 = 1/35 = 0.0285

P( The items both are baseball) = 0.0285

c) Without restriction the number of ways to select the 2 item from a bag = 15 C2 = 105

The number of ways to select one is biliard ball and other is an avacado = 2C1 * 7C1 = 14

P( one item is biliard ball and other is an avacado ) = 14/105 = 2/15 = 0.1333

P( one item is biliard ball and other is an avacado ) = 0.1333

d) Both items are food

The possible cases are

i) 1 apple and 1 avacado

ii) both are apple

iii) both are avacados

P( 1 apple and 1 avacados) = ( 3 *7 ) /105 = 21/105

P( both are apples) = 3C2 / 105 = 3/105

P( both are avacados) = 7C2 /105 =21/105

Hence

P( Both the items are food) = 21/ 105 + 3/105 + 21/105 = 45/105 = 3/7

P(Both the items are food) = 0.4285

Explanation / Answer

a. Pr( 1 avocado is removed and replaced and again another avocado is removed) = (7/15)*(7/15) = 0.2177

b. Pr( Both items are baseballs after not replacing the first removed item ) = 3C2 (Number of ways of choosing 2 baseballs from 3) / 15C2 (Number of ways of choosing 2 items from 15 )

= 3 / 105 = 1/35 = 0.0285

c. Pr( 1 billiard ball and 1 avocado is removed ) = 2C1 * 7C1 / 15C2 = 2/15 = 0.1333

d. Number of food items = 3+7 = 10

Pr( Both items are food ) = 10C2 / 15C2 = 3/7 = 0.4286

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

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