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sanjay mishra

asked 2017-03-05 10:11:17

An urn contains $6$ red, $4$ blue, $2$ green and $3$ yellow marbles. If four marbles are picked at random, what is the probability that at least one is blue?

$(a) \dfrac{5}{91}\quad(b) \dfrac{1}{35}\quad(c) \dfrac{1}{3}\quad(d) \dfrac{4}{105}\quad(e)$ None of these

$(a) \dfrac{5}{91}\quad(b) \dfrac{1}{35}\quad(c) \dfrac{1}{3}\quad(d) \dfrac{4}{105}\quad(e)$ None of these

Jay Patel

answered 2017-03-05 10:22:51

Answer is $\dfrac{69}{91}$ (none of the given choices)

Number of ways in which $4$ marbles can be picked

$=15\text{C}4$

Number of ways in which $4$ marbles can be picked where there is no blue marble

$=11\text{C}4=330$

Probability that there is no blue marble

$=\dfrac{11\text{C}4}{15\text{C}4}=\dfrac{22}{91}$

Probability that there is at least one blue marble

$=1-\dfrac{22}{91}=\dfrac{69}{91}$

Number of ways in which $4$ marbles can be picked

$=15\text{C}4$

Number of ways in which $4$ marbles can be picked where there is no blue marble

$=11\text{C}4=330$

Probability that there is no blue marble

$=\dfrac{11\text{C}4}{15\text{C}4}=\dfrac{22}{91}$

Probability that there is at least one blue marble

$=1-\dfrac{22}{91}=\dfrac{69}{91}$

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