Probability?

According to genetic theory, the blossom color in the second generationof a certain cross of sweet peas should be red or white in 3:1 ratio. That is, each plant has probability 3/4 of having redblossoms, and the blossomcolor of separate plants are independent.



(a) What is the probability that exactly 6 out of 8 of these plants have red blossoms?



(b) What is the mean number of red-blossomed plants when 80 plants of this type are grown from seeds?



(c) What is the probability of obtaining at least 50 red-blossomed plants whe 80 plants are grown frm seeds?
Probability?
Let X be the number of plants with red blossoms. X has the binomial distribution with n = 8 trials and success probability p = 0.75



In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, ..., n

P[X = x] = 0 for any other value of x.



The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.

Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.



X ~ Binomial( n , p )



the mean of the binomial distribution is n * p = 6

the variance of the binomial distribution is n * p * (1 - p) = 1.5

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.224745



The Probability Mass Function, PMF,

f(X) = P(X = x) is:



P( X = 0 ) = 1.525879e-05

P( X = 1 ) = 0.0003662109

P( X = 2 ) = 0.003845215

P( X = 3 ) = 0.02307129

P( X = 4 ) = 0.08651733

P( X = 5 ) = 0.2076416

P( X = 6 ) = 0.3114624 ← answer

P( X = 7 ) = 0.2669678

P( X = 8 ) = 0.1001129



== -- == -- == -- == -- ==



b)



80 * 0.75 = 60



== -- == -- == -- == -- ==



c)



To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p %26gt; 10 and n * (1-p) %26gt; 10.



Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker especially if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.



In this case you have:

n * p = 80 * 0.75 = 60 expected success

n * (1 - p) = 80 * 0.25 = 20 expected failures



We have checked and confirmed that there are enough expected successes and expected failures. Now we can move on to the rest of the work.



If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ2 = n * p * (1-p), and standard deviation σ



Xb ~ Binomial(n = 80 , p = 0.75 )

Xn ~ Normal( μ = 60 , σ2 = 15 )

Xn ~ Normal( μ = 60 , σ = 3.872983 )



I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.



The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.



P( Xb %26lt; x) ≈ P( Xn %26lt; (x - 0.5) )

P( Xb %26gt; x) ≈ P( Xn %26gt; (x + 0.5) )

P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5) )

P( Xb ≥ x) ≈ P( Xn ≥ (x - 0.5) )

P( Xb = x) ≈ P( (x - 0.5) %26lt; Xn %26lt; (x + 0.5) )

P( a ≤ Xb ≤ b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b + 0.5) )

P( a ≤ Xb %26lt; b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b - 0.5) )

P( a %26lt; Xb ≤ b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b + 0.5) )

P( a %26lt; Xb %26lt; b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b - 0.5) )



In the work that follows Xb has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.



Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - μ ) / σ



P( Xb ≥ 50 ) =



80

∑ P(Xb = x) = 0.995418

x = 50



≈ P( Xn ≥ 49.5 )

= P( Z ≥ ( 49.5 - 60 ) / 3.872983 )

= P( Z ≥ -2.711088 )

= 0.9966469
Reply:The Binomial distribution has the probability

function P(x=k)= nCr p^r (1-p)^(n-r)

k=0,1,2,.....,n

where nCr = n! / r! (n-r)!

a) n=8, x=6, p=3/4 or 0.75

0.3115



b)mean of the Binomial distribution is np = 80(3/4) =60



c)n=80, x=50 through 80, p=3/4 or 0.75

You may use Normal approximation, but let us directly evaluate it.

P(x %26gt;=50) =0.9954

I have included a source that would help you compute these probabilities. If you have access to a Binomial table, you may use it.