You come home from Brian’s Orchard with a big brown bag of apples: 23 Granny Smiths, 14 Honey Crisp and 31 Red Delicious. What is the probability you reach in and pull out: a. one of each? b. 4 Honey Crispsc. 2 G. Smiths & 1 R. Delicious Please help I am really confused and my teacher won’t respond to my email

a. Probability of Pulling one of each = 0.03175

b. Probability of Pulling 4 Honey Crisp = 0.001797

Probability of 2 G.Smith = 0.1144

Probability of 1 Red Delicious = 0.4559

Step-by-step explanation:

Given

Granny Smiths = 23

Honey Crisp = 14

Red Delicious = 31

Required

- Probability of Pulling out one of each

- Probability of Pulling out 4 Honey Crisp; 2 Granny Smiths; 1 Red Delicious

First, the total number of apple needs to be calculated.

Total = Granny Smiths + Honey Crisp + Red Delicious

Total = 23 + 14 + 31

Total = 68

Probability of Pulling 1 of each

= P(Granny Smiths) and P(Henry Ford) and P(Red Delicious)

- Granny Smiths;

This is calculated by dividing number of Granny Smiths apples by total number of apples.

Probability = 23/68

Similarly,

Probability of Pulling Honey Crisp= Number of Honey Crisp divided by total

Probability = 14/68

Probability = 7/34

Probability of Pulling Red Delicious = Number of Red Delicious divided by total

Probability = 31/68

So, Probability of Pulling 1 of each = 23/68 * 7/34 * 31/68

Probability = 4991/157216

Probability = 0.03175

Probability of Pulling out 4 Honey Crisp;

= P(Honey) * P(Honey) * P(Honey) * P(Honey)

= (P(Honey))⁴

= (7/34)⁴

= 2401/1336336

= 0.001797

Probability of Pulling 2 Granny Smiths;

= P(Granny) * P(Granny)

= (P(Granny))²

= (23/68)²

= 529/4624

= 0.1144

Probability of 1 Red Delicious

= number of red delicious divided by total

= 31/68

= 0.4559

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