The quality control manager of Marilyn's Cookies is inspecting a batch of chocolate-chip cookies that has just been baked. If the production process is in control, the mean number of chip parts per cookie is 6.5. Complete parts (a) through (c).
a. What is the probability that in any particular cookie being inspected less than five chip parts will be found?
b. What is the probability that in any particular cookie being inspected exactly five chip parts will be found?c. What is the probability that in any particular cookie being inspected five or more chip parts will be found?d. What is the probability that in any particular cookie being inspected either four or five chip parts will be found?

(a) P(X<5) = 0.2237

(b) P(X=5) = 0.1454

(c) P(X≥5) = 0.7763

(d) P(X=either 4 or 5) = 0.2572

Step-by-step explanation:

We will consider this question as a Poisson process and use the following formula to determine the answers:

P(X=x) = λˣe^-λ / x!

Where λ = Mean number of chip parts per cookie

           x = Number of chip parts found

(a) For calculating the probability of less than 5 chip parts being found, we need to consider all the values less than 5 i.e. 0,1,2,3,4 and calculate the probability for these values.

P(X < 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)

                  = /0! + /1! + /2! + /3! + /4!

     P(X<5) = 0.0015 + 0.0098 + 0.0318 + 0.0688 + 0.1118

     P(X<5) = 0.2237

The probability that in any particular cookie being inspected, less than 5 chip parts will be found is 0.2237

(b) P(X=5) = /5!

    P(X=5) = 0.1454

The probability that in any particular cookie being inspected, exactly 5 chip parts will be found is 0.1454.

(c) To calculate the probability of more than or equal to 5 chip parts being found, we will consider the probability P(X<5) which we have already found in part (a) and subtract it from the total probability (i.e. 1) to get P(X≥5).

P(X≥5) = 1 - P(X<5)

                = 1 - 0.2237

    P(X≥5) = 0.7763

The probability that in any particular cookie being inspected 5 or more chip parts will be found is 0.7763.

(d) To calculate the probability that either 4 or 5 chip parts will be found, we must add P(X=4) and P(X=5). We know the value of P(X=5) from part (b) and the value of P(X=4) from part (a). Hence,

P(X=either 4 or 5) = P(X=4) + P(X=5)

                              = 0.1118 + 0.1454

P(X=either 4 or 5) = 0.2572

The probability that in any particular cookie being inspected, either 4 or 5 chip parts will be found is 0.2572.

based on context clues you gave i'm assuming the answer would be c.

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b