A simple random sample of size five is selected from 52 cards. Find the probabil

Question

A simple random sample of size five is selected from 52 cards. Find the probability of each of the following events. (a) The five-card hand contains two aces and two 7's and a card other than ace and 7. (b) The five-card hand contains three kings and the other two cards are not kings and of different kinds. Each of the machines A and B in an electronics fabrication plant produces a single batch of 50 electrical components per hour. Let E_1 denote the event that, in any given hour, machine A produces a batch with no defective components, and E_2 denote the corresponding event for machine B. The probabilities of E_1, E_2, and E_1 intersection E_2 are 0.95, 0.92 and 0.88, respectively. Express each of the following events as set operations on E_1 and E_2, and find their probabilities. (a) In any given hour, only machine A produces a batch with no defects. (b) In any given hour, only machine B produces a batch with no defects. (c) In any given hour, exactly one machine produces a batch with no defects. (d) In any given hour, at least one machine produces a batch with no defects.

Explanation / Answer

Problem 2

a) We have to choose 5 cards from 52 cards, we can do it in 52C5 ways

Now we know that a pack of 52 cards has 4 set of 13 cards of different color

so we can choose 2 Aces from 4 cards

We can choose 2 "7" cards from 4 cards

For choosing a card different from ace and "7" card we choose the card from the remainig 44 cards (52-8)

Therefore the Answer is (4C2*4C2*44C1)/52C5

b) Now we have to choose 3 Kings

We only have 4 diffrent kings in the pack of 52

This can be done in 4C3 ways

Now for the remainig 2 cards we can choose them from the other 48 cards ie 52-4 king cards

So the answer is( 4C3*48C2)/52C5

Problem 3

Given to us

E1=P(A)=0.95

E2=P(B)=0.92

not E1=P(A')=0.05

not E2=P(B')=0.08

a) Only A produces without defects ie P(A U B') is being asked

=1-P(B)+P(A intersection B)

=1-0.92+0.88

=0.96

b) Only B produces without defects is P(A' U B)

=1-P(A)+ P( A intersection B)

=1-0.95+0.88

=0.93

c) Exactly one produces with no defect so either A can produce without defects or B can produce without defects

P(A intersection B') + P( A' intersection B )

=P (A U B)-P(A intersection B)

=P(A) +P (B)-P(A intersection B)-P(A intersection B)

=0.95+0.92-0.88-0.88

=0.11

d) Atleast 1 means 1 or 2

so we need to calculate P(A U B)=P(A)+P(B)-P(A intersection B)

=0.95+0.92-0.88

=0.99

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