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A stationery bag contains five pencils, two erasers and five pens. If three stationeries are randomly selected from the bag (with replacement), what is the probability that

a) one pencil, one eraser and one pen b) the three item selected either all pencils or all pens c) none of the pencils is selected

Please help me to solve this question

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  • $\begingroup$ Hint for a) The probability that a pencil, a eraser and a pen are selected in that order (with replacement) is $\frac{5}{12}\cdot \frac{2}{12}\cdot \frac{5}{12}$. But there are other possible orders like (eraser, pencil, pen), (pencil, pen, eraser), ... In total $3!=6$. $\endgroup$ May 7, 2017 at 18:30

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There is $\frac{5}{12}$ chance of choosing a pencil or a pen if only one item is selected. Similarly there is $\frac{2}{12}$ chance of choosing eraser.

Starting with (a):

Suppose I want to choose those elements IN ORDER. Then I can use the product rule to determine that there is $\frac{5*2*5}{12*12*12}$ chance of choosing a pencil, an eraser and a pen in this order. However the order was not required so I have to consider in how many different orders 3 items can appear - the answer is 3! - and then multiply this number by the previous one.

For (b) you can use multiplication rule again.

For (c) you can note that probability of not selecting a pencil i 1 - probability of selecting a pencil and use multiplication rule once again.

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  • $\begingroup$ could you show me the steps to solve the (c) ? $\endgroup$ May 7, 2017 at 19:14

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