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The word PROBABILITY is randomly arranged in a row. Q find the probability that $Y$ is not in the last position and the two $B$'s are not consecutive.

Let $B = \{ \text{Event that two B's are together}\}$ and $Y$ be the event of last Y.

We want $P(\overline{B} \cap \overline{Y}) = 1 - P(B \cup Y) = 1 - [P(B) + P(Y) - P(BY)]$

I calculated, $P(B) = 0.182$, $P(Y) = 0.091$, $P(BY) = 0.018$, thus $1 - P(B \cup Y) = 0.709$, but the answer key says: $0.745$?

What am I doing wrong?

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    Why is Y the event of consecutive 'Y's? Isn't it meant to be the event of a final 'Y'? – 2017-01-31
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    @ConMan, thats what I meant, sorry – 2017-01-31
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    For what it's worth, I get the same value you did. Just to point out: the official answer differs from the one you got by $.036=2\times P(B\cap Y)$...so perhaps they added where they meant to subtract? – 2017-01-31
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    @lulu, so which answer is correct? – 2017-01-31
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    As I say, I get the same answer you get so I believe that one to be correct ($\frac {39}{55}$ to be precise). I mentioned the $.036$ as a possible source of error on your text's part. (Of course, it is perfectly possible that you and I are making an error in common). – 2017-01-31
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    $1-(0.182+0.091-0.018)=0.745$ – 2017-01-31

1 Answers 1

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The total space is $11!$ equal probable arrangements of letters.

The event of $B$ is $10!\,2!$ equal probable arrangements of letters. $~P(B)=\frac{10!\,2!}{11!}\approx 0.182$

The event of $Y$ is $10!$ equal probable arrangements of letters. $~~~~P(Y)= \frac{10!}{11!}\approx 0.091$

The event of $B\cap Y$ is $9!2!$ equal probable arrangements of letters. $P(B\cap Y)= \frac{9!\,2!}{11!}\approx 0.018$

$$P(\overline B\cap \overline Y)=1-\tfrac{10!\,2!+10!-9!\,2!}{11!} \approx 0.745$$

So you were correct right up to putting it together, where you apparently made a basic calculator entry error due to signs in brackets.

$1-\big(0.182+0.091-0.018\big)~=~1-0.182-0.091+0.018~=~0.745$