Abox has 14 camera of which 6 are refurbished and 8 are new. if four of these 14 cameras are selected at random without replacement, what is the probability that (i) one new camera will be selected? (ii) at most one new camera will be selected?

160/1001, 175/1001

Step-by-step explanation:

i) There are:

₈C₁ ways to choose 1 new camera from 8 new cameras

₆C₃ ways to choose 3 refurbished cameras from 8 refurbished cameras

₁₄C₄ ways to choose 4 cameras from 14 cameras

The probability is:

P = ₈C₁ ₆C₃ / ₁₄C₄

P = 8×20 / 1001

P = 160 / 1001

P ≈ 0.160

ii) At most one new camera means either one new camera or no new cameras.  We already found the probability of one new camera.  The probability of no new cameras is the same as the probability of choosing 4 refurbished cameras:

P = ₆C₄ / ₁₄C₄

P = 15 / 1001

So the total probability is:

P = 160/1001 + 15/1001

P = 175/1001

P ≈ 0.175

the difference between a number and 5 is -2.

step-by-step explanation:

notice that the equation you are given has -2 (negative 2) on the right side, not 2. when you subtract 5 from a number n, you are finding the difference between the number and 5.

the difference between a number and 5 is -2.

answer: 2 and 1

step-by-step explanation: got this question