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:
Step-by-step explanation: You can multiply 11 by 15 to get 165. If that is too difficult, you can multiply 11 by 5 and 11 by 10 and add them. 11*5 is 55 and 11*10 is 110. 110 + 55 = 165.
Answer: about 66.6% that he will not pick a PB one
Step-by-step explanation: Put both of them together and you get 15. 15 divede by 3 is 5. Two 5's is 10, which is about nine, so it would be 2/3 chance of getting an oatmeal cookie, which is the opposite of getting a pb one, 2/3 in a percentage is 66.6%
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