42
Find the probability that a number from 1 to 300 is either divisible by 3 or 7?
- A$$\frac{37}{75}$$
- B$$\frac{32}{75}$$
- C$$\frac{36}{75}$$
- D$$\frac{28}{75}$$
Answer not set
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0
CCommunity
Solution
Numbers divisible by 3 = {3, 6, 9, ...... 300}
Numbers divisible by 7 = {7,14, ........ 294}
Numbers divisible by both 3 and 7 (i.e. multiples of 21) = {21, 42,.... 294}
Required probability = $\frac{100 + 42 - 14}{300} = \frac{128}{300} = \frac{32}{75}$
Numbers divisible by 3 = {3, 6, 9, ...... 300}
Numbers divisible by 7 = {7,14, ........ 294}
Numbers divisible by both 3 and 7 (i.e. multiples of 21) = {21, 42,.... 294}
Required probability = $\frac{100 + 42 - 14}{300} = \frac{128}{300} = \frac{32}{75}$
0
CCommunity
Solution
Numbers divisible by 3 = {3, 6, 9, ...... 300}
Numbers divisible by 7 = {7,14, ........ 294}
Numbers divisible by both 3 and 7 (i.e. multiples of 21) = {21, 42,.... 294}
Required probability = $\frac{100 + 42 - 14}{300} = \frac{128}{300} = \frac{32}{75}$
Numbers divisible by 3 = {3, 6, 9, ...... 300}
Numbers divisible by 7 = {7,14, ........ 294}
Numbers divisible by both 3 and 7 (i.e. multiples of 21) = {21, 42,.... 294}
Required probability = $\frac{100 + 42 - 14}{300} = \frac{128}{300} = \frac{32}{75}$