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In a box carrying one dozen of oranges one third have become bad.If 3 oranges taken out from the box random ,what is the probability that at least one orange out of the 3 oranges picked up is good ?
- A1/55
- B54/55
- C45/55
- D3/55
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0
CCommunity
Solution
Total number of oranges in the box = 12
Number of ways of selecting 3 oranges out of 12 oranges, n(S) = $C^{12}_3$
= $\frac{12 \times 11 \times 10}{1 \times 2 \times 3} = 220$
Number of oranges which became bad = $\frac{12}{3}=4$
Number of ways of selecting 3 oranges out of 4 bad oranges = $C^4_3 = C^4_1 = 4$
Number of desired selection of oranges, n(E) = 220 - 4 = 216
$\therefore$ $P(E) = \frac{n(E)}{n(S)}$
= $\frac{216}{220}= \frac{54}{55}$
=> Ans - (B)
Total number of oranges in the box = 12
Number of ways of selecting 3 oranges out of 12 oranges, n(S) = $C^{12}_3$
= $\frac{12 \times 11 \times 10}{1 \times 2 \times 3} = 220$
Number of oranges which became bad = $\frac{12}{3}=4$
Number of ways of selecting 3 oranges out of 4 bad oranges = $C^4_3 = C^4_1 = 4$
Number of desired selection of oranges, n(E) = 220 - 4 = 216
$\therefore$ $P(E) = \frac{n(E)}{n(S)}$
= $\frac{216}{220}= \frac{54}{55}$
=> Ans - (B)
0
CCommunity
Solution
Total number of oranges in the box = 12
Number of ways of selecting 3 oranges out of 12 oranges, n(S) = $C^{12}_3$
= $\frac{12 \times 11 \times 10}{1 \times 2 \times 3} = 220$
Number of oranges which became bad = $\frac{12}{3}=4$
Number of ways of selecting 3 oranges out of 4 bad oranges = $C^4_3 = C^4_1 = 4$
Number of desired selection of oranges, n(E) = 220 - 4 = 216
$\therefore$ $P(E) = \frac{n(E)}{n(S)}$
= $\frac{216}{220}= \frac{54}{55}$
=> Ans - (B)
Total number of oranges in the box = 12
Number of ways of selecting 3 oranges out of 12 oranges, n(S) = $C^{12}_3$
= $\frac{12 \times 11 \times 10}{1 \times 2 \times 3} = 220$
Number of oranges which became bad = $\frac{12}{3}=4$
Number of ways of selecting 3 oranges out of 4 bad oranges = $C^4_3 = C^4_1 = 4$
Number of desired selection of oranges, n(E) = 220 - 4 = 216
$\therefore$ $P(E) = \frac{n(E)}{n(S)}$
= $\frac{216}{220}= \frac{54}{55}$
=> Ans - (B)