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PART - 2 CH - 14 Probability — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsPART - 2 CH - 14 Probability5 Marks
Question
$A$ and $B$ are throwing two dice one after the other. If before $B$ throwing a 7, $A$ throws 6 then $A$ wins. If $B$ throws a 7 before $A$ throws a 6 then $B$ wins. If $A$ starts throwing then prove that the probability of winning of $A$ is $\frac{30}{61}.$
✓

Answer

Favourable outcomes for obtaining a total of 6 on two dice $=(1,5),(2,4),(3,3),(4,2)$ and $(5,1)$
Probability of getting a sum of $6=\frac{5}{36}$
Probability of not getting a sum of $6=1-\frac{5}{36}=\frac{31}{36}$
In the same way, favourable outcomes for obtaining a sum of 7 are $(1,6),(2,5),(3,4),(4,3),(5,2)$ and $(6,1)$
Probability of getting a sum of $7=\frac{6}{36}=\frac{1}{6}$
Probability of not getting a sum of $6=1-\frac{1}{6}=\frac{5}{6}$
A start then the following possibilities of winning of A are
(i) A gets a 6 in first throw whose probability is $\frac{5}{36}$
(ii) A does not get 6 in first throw, B does not get 7 in second throw and A gets 6 in third throw whose probability will be $\frac{31}{36} \times \frac{5}{6} \times \frac{5}{36}$
(iii) A does not get a 6 in first throw, B does not get 7 in second throw, A does not get 6 in third throw, B does not get a 7 in fourth throw and A gets a 6 in fifth throw whose probability will be
$\left(\frac{31}{36}\right) \times\left(\frac{5}{6}\right) \times\left(\frac{31}{36}\right) \times\left(\frac{5}{6}\right) \times \frac{5}{36}$
$=\left\{\left(\frac{31}{36}\right) \times\left(\frac{5}{6}\right)\right\}^2 \times \frac{5}{36}$
If we assume that A throws dice infinite times to win and since all these events are mutually exclusive, then by addition theorem of probability, probability of winning of $A$ is
$=\frac{5}{36}+\frac{5}{36}\left\{\left(\frac{31}{36}\right) \times\left(\frac{5}{6}\right)\right\}+\frac{5}{36}\left\{\left(\frac{31}{36}\right) \times\left(\frac{5}{6}\right)\right\}^2+\ldots$
The above series is infinite series of geometric progression. Hence sum of infinite terms
$\begin{aligned}=S_{\infty} & =\frac{a}{1-r} \\& =\frac{\frac{5}{36}}{1-\left\{\left(\frac{31}{36}\right)\times\left(\frac{5}{6}\right)\right\}}=\frac{5}{36} \times \frac{36 \times 6}{61} \\& =\frac{30}{61}\end{aligned}$

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