Question

Can someone help me with these questions please

Thank you

6. It is common in many industrial areas to use a filling machine to ill boxes full of product This occurs in the food industry as well as other areas in which the product is used in the home, for example, detergent. These machines are not perfect, and indeed they may A, fill to specification, B, underfill, and C, overfill. Generally the practice of underfilling is that which one hopes to avoid. Let P(C) = 0.034 while P(A) = 0.960. (a) What is the probability that the box is underfilled, P(B)? (b) What is the probability that the machine does not overfill? (c) What is the probability that the machine either overfills or underfills? 7. In how many different ways can a true-false test consisting of 12 questions be answered? 8. Four married couples have bought S seats in the same row for a concert. In how man different ways can they be seated (a) with no restrictions? (b) if each couple is to sit together? (c) if all the husbands sit together to the right of all the wives? 9. A machinist produces 22 items during a shift. Three of the 22 items are defective and the rest are not defective. In how many different orders can the 22 items be arranged if all the defective items are considered identical and all the non-defective items are identical of a different class?

Answer 1

Solution : ( 7 )

On each of the 12 questions can be answered in 2 ways.

So, each of 12 operations can result in 2 different ways.

Hence, in this case we have r = 12 and n_i = 2, i = 1, 2, 3, 4, . . . , 12.

n₁ = 2, n₂ = 2, n₃ = 2, n₄ = 2, n₅ = 2, n₆ = 2, n₇ = 2, n₈ = 2, n₉ = 2, n₁₀ = 2, n₁₁ = 2, n₁₂ = 2.

Now,

n_{1 *} n_{2 *} n_{3 *} n_{4 *} n_{5 *} n_{6 *} n_{7 *} n_{8 *} n_{9 *} n_{10 *} n_{11 *} n₁₂

= (2) * (2) * (2) * (2) * (2) * (2) * (2) * (2) * (2) * (2) * (2) * (2)

= 2¹²

= 4096 ways

So, a true - false test consisting of 12 questions can be answered in 4096 different ways.

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Solution : ( 8 )

( a ) Since there is no restriction, the number of ways in which 4 couples can occupy the seats

= 8!

= 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

= 40320

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( b ) 4! ways of arranging couples, each couple can interchanging seats, so total 2⁴ * 4! = 16 * 4 * 3 * 2 * 1 = 384 ways.

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( c ) Wives can occupy seats in 4! and husbands can occupy seats in 4! .

Hence, all the husbands sit together to the right of all the wives = 4! * 4! = 24 * 24 = 576 ways.

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Solution : ( 9 )

Number of ways of arranging 3 defective items and 19 nondefective items is :

= 22C₁₉

_{= 22! / ( 19! * ( 22 - 19)! )}

_{= 22! / ( 19! * 3! )}

_{= ( 22 * 21 * 20 * 19! ) / ( 19! * 3! )}

_{= ( 22 * 21 * 20 ) / ( 3! )}

_{= ( 22 * 21 * 20 ) / ( 3 * 2 * 1 )}

_{= 1540}