Question
A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If 5 balls are picked up at random, what is the probability that at least one is blue?
Answer: Option A
Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 5 balls out of 9 non-blue balls.
$$ = {}^9\mathop C\nolimits_5 = {}^9\mathop C\nolimits_{\left( {9 - 5} \right)} = {}^9\mathop C\nolimits_4 $$ $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$ = 126
And,
n(S) = $${}^{15}\mathop C\nolimits_5 = $$ $$\frac{{15 \times 14 \times 13 \times 12 \times 11}}{{5 \times 4 \times 3 \times 2 \times 1}}$$ = 3003
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = $$ $$\frac{{126}}{{3003}} = \frac{6}{{143}}$$
∴ Required Probability
$$\eqalign{
& = \left( {1 - \frac{6}{{143}}} \right) \cr
& = \frac{{137}}{{143}} \cr} $$
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Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 5 balls out of 9 non-blue balls.
$$ = {}^9\mathop C\nolimits_5 = {}^9\mathop C\nolimits_{\left( {9 - 5} \right)} = {}^9\mathop C\nolimits_4 $$ $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$ = 126
And,
n(S) = $${}^{15}\mathop C\nolimits_5 = $$ $$\frac{{15 \times 14 \times 13 \times 12 \times 11}}{{5 \times 4 \times 3 \times 2 \times 1}}$$ = 3003
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = $$ $$\frac{{126}}{{3003}} = \frac{6}{{143}}$$
∴ Required Probability
$$\eqalign{
& = \left( {1 - \frac{6}{{143}}} \right) \cr
& = \frac{{137}}{{143}} \cr} $$
Was this answer helpful ?
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