\(A\) and \(B\) throw a pair of dice. \(A\) wins if he throws 6 before \(B\) throws 7
and \(B\) wins if hethrows 7 before \(A\) throws 6 . If \(A\) begins, what is his
chance of winning?
(a) \(30 / 61\)
[MNR 1995]
(c) \(61 / 30\)
(1)
olution
(a) Let \(E_{1}\) denote the event of \(A\) 's throwing 6 and \(E_{2}\) the event of
\(B\) 's throwing 7 with a pair of dice.
Then \(\bar{E}_{1}, \bar{E}_{2}\) are the complementary events. There are five
ways of obtaining 6, namely, \((1,5),(2,4),(3,3),(4,2),(5,1)\) and similarly
there are six ways of getting 7 , namely, \((1,6)\),
\((2,5),(3,4),(4,3),(5,2),(6,1)\)
$$
\therefore P\left(E_{1}\right)=\frac{5}{36}
$$
and \(P\left(\bar{E}_{1}\right)=1-\frac{5}{36}=\frac{31}{36}\)
$$
P\left(E_{2}\right)=\frac{6}{36}=\frac{1}{6}
$$
and \(P\left(\bar{E}_{2}\right)=1-\frac{1}{6}=\frac{5}{6}\)
It is given that \(A\) starts the game and he will win in the following mutually
exclusive ways.
(i) \(E_{1}\) happens, i.e., \(A\) wins at the first draw.
(ii) \(\vec{E}_{1} \cap \bar{E}_{2} \cap E_{1}\) happens, i.e., \(A\) wins at the
third draw when both \(A\) and \(B\) fail at \(1^{s t}\) and \(2^{\text {nd }}\) draw.
(iii) \(\bar{E}_{1} \cap \bar{E}_{2} \cap \bar{E}_{1} \cap \bar{E}_{2} \cap
E_{1}\) happens, i.e., \(A\) wins at the \(5^{\text {th }}\) draw when both \(A\) and
\(B\) fail at \(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}\) and \(4^{\text
{th }}\) draw and so on \(\ldots\) Hence the required probability of \(A\) winning
say \(P(A)\) is given by \(P(A)=P(\mathrm{i})+P(\mathrm{ii})+P(\mathrm{iii})\)
\(+\ldots\)throws 7 before \(A\) throws 6 . If \(A\) begins, what is his chance of
winning?
(a) \(30 / 61\)
[MNR 1995]
(c) \(61 / 30\)
(1)
olution
(a) Let \(E_{1}\) denote the event of \(A\) 's throwing 6 and \(E_{2}\) the event of
\(B\) 's throwing 7 with a pair of dice.
Then \(\bar{E}_{1}, \bar{E}_{2}\) are the complementary events. There are five
ways of obtaining 6, namely, \((1,5),(2,4),(3,3),(4,2),(5,1)\) and similarly
there are six ways of getting 7 , namely, \((1,6)\),
\((2,5),(3,4),(4,3),(5,2),(6,1)\)
$$
\therefore P\left(E_{1}\right)=\frac{5}{36}
$$
and \(P\left(\bar{E}_{1}\right)=1-\frac{5}{36}=\frac{31}{36}\)
$$
P\left(E_{2}\right)=\frac{6}{36}=\frac{1}{6}
$$
and \(P\left(\bar{E}_{2}\right)=1-\frac{1}{6}=\frac{5}{6}\)
It is given that \(A\) starts the game and he will win in the following mutually
exclusive ways.
(i) \(E_{1}\) happens, i.e., \(A\) wins at the first draw.
(ii) \(\vec{E}_{1} \cap \bar{E}_{2} \cap E_{1}\) happens, i.e., \(A\) wins at the
third draw when both \(A\) and \(B\) fail at \(1^{s t}\) and \(2^{\text {nd }}\) draw.
(iii) \(\bar{E}_{1} \cap \bar{E}_{2} \cap \bar{E}_{1} \cap \bar{E}_{2} \cap
E_{1}\) happens, i.e., \(A\) wins at the \(5^{\text {th }}\) draw when both \(A\) and
\(B\) fail at \(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}\) and \(4^{\text
{th }}\) draw and so on \(\ldots\) Hence the required probability of \(A\) winning
say \(P(A)\) is given by \(P(A)=P(\mathrm{i})+P(\mathrm{ii})+P(\mathrm{iii})\)
\(+\ldots\)
Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine 
