Deers eat ornamental cabbage plants There are 10 plants in t

Deers eat ornamental cabbage plants. There are 10 plants in the outdoor transplant area, which will be ready to plant in 10 days. The probability of the deer eating the plant prior to transplanting is 0.60. With X being a binomail probability distribution what is the numer of trials (n), probability of successes (p) and probability of failures (q) respectively. What is the probability that there will be at least 2 of the 10 ornamental plants to transplant and how many can be expected to thrive after transplanting.

Solution

The probability that a plant is eaten is 0.60.

Thus, the probability that it survives is 1 - 0.60 = 0.40.

THus,

p = 0.40 here,

where a success is when it is not eaten.

Note that

P(at least 2) = 1 - P(0) - P(1)

Now, we use a binomial probability distribution.

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    10      
p = the probability of a success =    0.4      
x = the number of successes =    0      
          
Thus, the probability is          
          
P(0) =    0.006046618

The same way,

P(1) = 0.040310784

Thus,

P(at least 2) = 0.953642598 [ANSWER]

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The number of plants expected to thrive is

mean = n p = 10*0.40 = 4 [ANSWER]