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there are 11 sweets in a box four are soft centred and seven hard centred sweets

two sweets are selected at random

a)calculate the probability that both sweets are hard centred,

b) one sweet is soft centred and one sweet is hard centred this what I done

a) both sweets are hard centred is 4/11 * 7/11=28/121 is this right? I am not sure for part b

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    are sweets selected together or separately? – 2017-02-05
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    @user362325 they are selected separately – 2017-02-05
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    This is drawing w/o replacement, number of sweets in box will keep on reducing. Have you studied the hypergeometric distribution ? – 2017-02-05
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    yes I understand @trueblueanil, no I have not studied the hypergeometric distribution – 2017-02-05
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    You post many questions with practically the same title which says nothing about the specific question. Also, the "what I did" part seems like you just take some numbers from the problem and make some wild guesses to combine them in a formula. You should explain _why_ each particular number occurs so we can see you actually thought about the problem. – 2017-02-28

2 Answers 2

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Notice that if you firstly select a sweet with hard center, then there are only 10 sweets left and 6 sweets of them are hard centered.

So the probability of question (a) is $\frac{7}{11}*\frac{6}{10}$.

For question (b), you should think about which kind of sweets is selected first.

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For b): the event you are after (one soft, one hard) can happen two ways: SH or HS. In other words, you want $$ P(SH \cup HS) = P(SH) + P(HS) - P(SH \cap HS) $$ The events are disjoint so $P(SH \cap HS) = 0$. Next you have $$ P(SH) = \frac{4}{11} \cdot \frac{7}{10} $$ Can you figure out $P(HS)$?

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    Wouldn't that just be the complement? – 2017-02-05