Jose has a jar of 43 candy pieces 17 are red 11 are blue 8 a

Jose has a jar of 43 candy pieces. 17 are red, 11 are blue, 8 are green and 7 are yellow. He randomly chooses eight pieces of candy. Calculate the probability for each given event.

1.No red pieces are choosen

2.Three pieces are green and two are yellow

3.Five of the pieces are red or blue

4.Six pieces are blue or six pieces are red

5.An equal number of each color are choosen

Solution

1. No red pieces are choosen

Probability of picking 8 candies from 43 candies (Total number of events) : ⁴³C₈ (=145008513)

Probability of not picking all pieces as red : ²⁶C₈ (=1562275)

Therefore, probability that No red pieces are choosen : ²⁶C₈ / ⁴³C₈

2.Three pieces are green and two are yellow

Probability of picking 8 candies from 43 candies (Total number of events) : ⁴³C₈ (=145008513)

Probability that Three pieces are green : ⁸C₃ (=56)

Probability that two are yellow : ⁷C₂ (=21)

Probability of picking remaining 3 candies from 28 total remaining candies : ²⁸C₃ (=3276)

Therefore, probability that Three pieces are green and two are yellow = ( ⁸C_3* ⁷C_2* ²⁸C_{3 ) /} ⁴³C₈

3) Five of the pieces are red or blue

Probability of picking 8 candies from 43 candies (Total number of events) : ⁴³C₈ (=145008513)

Probability of Five of the pieces are red : ¹⁷C₅ (=6188)

In this case, probability of picking remaining 3 candies from total remaining 26 candies : ²⁶C₃ (=2600)

Probability of Five of the pieces are blue : ¹¹C₅ (=462)

In this case, probability of picking remaining 3 candies from total remaining 32 candies : ³²C₃ (=4960)

Therefore, probability of Five of the pieces are red or blue : ( ¹⁷C_5*²⁶C₃ + ¹¹C_5*³²C₃ ) / ⁴³C₈

4) Six pieces are blue or six pieces are red

Probability of picking 8 candies from 43 candies (Total number of events) : ⁴³C₈ (=145008513)

Probability of Six of the pieces are red : ¹⁷C₆ (=12376)

In this case, probability of picking remaining 2 candies from total remaining 26 candies : ²⁶C₂ (=325)

Probability of Six of the pieces are blue : ¹¹C₆ (=462)

In this case, probability of picking remaining 2 candies from total remaining 32 candies : ³²C₂ (=496)

Therefore, probability of Five of the pieces are red or blue : ( ¹⁷C_6*²⁶C₂+ ¹¹C_6*³²C₂ ) / ⁴³C₈

5) An equal number of each color are choosen

Probability of picking 8 candies from 43 candies (Total number of events) : ⁴³C₈ (=145008513)

Picking equal number of each color means picking 2 candies each of every color

Therefore,

Probability of picking 2 blue candies : ¹¹C₂ (=55)

Probability of picking 2 red candies : ¹⁷C₂ (=136)

Probability of picking 2 green candies : ⁸C₂ (=28)

Probability of picking 2 yellow candies : ⁷C₂ (=21)

Therefore, probability An equal number of each color are choosen : (¹¹C_2*¹⁷C_2*⁸C_2*⁷C₂)/ ⁴³C₈