The seeds of the garden pea are either yellow or green. A certain cross between pea plants produces progeny in the ratio 3 yellow:1 green. If four randomly chosen progeny of such a cross are examined, what is the probability that:
a)3 are yellow and 1 is green
b)all 4 are yellow
c)all 4 are the same color
P=3/4
q=1/4
n=4
a) Ah, the classic yellow and green pea drama! To calculate the probability of 3 yellow and 1 green, we need to consider the ratio given. Since the ratio is 3 yellow to 1 green, there are a total of 4 possible outcomes, 3 of which are yellow and 1 of which is green. So, the probability would be 3/4. Just like a traffic light, it's mostly yellow with a hint of green!
b) Now, let's dive into the all-yellow scenario! If we assume that the first progeny chosen is yellow, then the probability of the second one also being yellow would be 2/3, the third one being yellow would be 1/2, and the fourth one being yellow would be 1/1 (since all the others are yellow). To find the total probability, we multiply these probabilities together: (1/1) x (1/2) x (2/3) x (3/4) = 1/4. So, the chances of all four being yellow are like finding a pot of gold at the end of the pea rainbow, 1 out of 4!
c) The final scenario is when all four progeny are the same color. Let's consider two possibilities: all yellow or all green. From previous calculations, we already know that the probability of all four being yellow is 1/4. Similarly, the probability of all four being green would also be 1/4. If we add these two probabilities together, we get 1/4 + 1/4 = 2/4 = 1/2. So, the chances of all four being the same color are like flipping a pea coin, 1 out of 2!
To calculate the probabilities, we first need to determine the number of possible outcomes. In this case, each progeny has two possible outcomes: yellow or green. So, there are 2^4 = 16 possible outcomes for four progeny.
a) To find the probability that 3 are yellow and 1 is green, we need to determine how many ways this outcome can occur. To do this, we can use the binomial coefficient formula, also known as the combination formula: nCr = n! / (r!(n-r)!).
In this case, n = 4 (the total number of progeny) and r = 3 (the number of yellow progeny). So, we have 4C3 = 4! / (3!(4-3)!) = 4! / (3! * 1!) = 4.
Therefore, there are 4 ways to have 3 yellow and 1 green progeny. The probability can be calculated as 4/16 = 1/4.
b) To find the probability that all 4 progeny are yellow, we need to determine the number of ways this outcome can occur. Using the same formula, we have 4C4 = 4! / (4!(4-4)!) = 1.
Therefore, there is 1 way to have all 4 progeny yellow, and the probability is 1/16.
c) To find the probability that all 4 progeny are the same color, we need to consider two cases: all yellow or all green progeny.
For all yellow progeny, we have 4C4 = 1 way (as calculated above). For all green progeny, we also have 4C4 = 1 way. So, there are a total of 2 ways to have all 4 progeny the same color.
The probability can be calculated as 2/16 = 1/8.
Therefore, the probabilities are:
a) 3 are yellow and 1 is green: 1/4
b) all 4 are yellow: 1/16
c) all 4 are the same color: 1/8
p yellow = 3/4
p not yellow = 1/4
n = number of trials = 4
from pascal's triangle or binomial distribution table or calculation for n = 4
P(0y) = 1 (3/4)^0 (1/4)^4
P(1y) = 4 (3/4)^1 (1/4)^3
P(2y) = 6 (3/4)^2 (1/4)^2
P(3y) = 4 (3/4)^3 (1/4)^1
P(4y) = 1 (3/4)^4 (1/4)^0
I think you can get it from there
a.3/16
b.9/16
c.1