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Detailed Solution

247. c: Let E5 = Event of getting a total of at least 10.
E5 = { (4,6), (5,5), (5,6), (6,4), (6,5), (6,6), }
n(E5) = 6
P(E5) = n(E5)/n(S) = 6/36 = 1/6

248. c: Let E3 = Event of getting ‘3’ non – red balls. So now we have to choose all the three balls from 4 white and 8 blue balls.
Total number of ways :
n(E3) = 12C3 = 12/ (3x 9) = (12x11x10) / (3x2x1) = 220
P(E3) = n(E3) / n(S) = 220 / 816 = 55/204

249. c: Let E1 = Event that both the tickets have prime numbers Prime numbers between ‘1’ to ‘50’ are : 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47.
Total Numbers = 15.
We have to select ‘2’ numbers from these 15 numbers.
n(E1) = 15C2 = 15? / (2? x 13?) = (15x47) / 2 = 105
P(E1) = n(E1) / n(S) = 105/1225 = 21/245

250.

251. b: The correct answer was 1/18. The probability of flipping heads is 1/2. The probability of rolling a composite number (4 and 6) is 1/3. The probability of spinning a 2 is 1/3. Multiply: 1/2 * 1/3 * 1/3 is equal to 1/18.

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