81. An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark ‘Y’. A ball is drawn at random from the urn, its mark noted down and it is replaced. If 6 balls are drawn in this way, find the probability that:

(i) All will bear ‘X’ mark.

(ii) Not more than 2 will bear ‘Y’ mark.

(iii) At least one ball will bear ‘Y’ mark.

(iv) The number of balls with ‘X’ mark and ‘Y’ mark will be equal.

2 Views | Posted 4 months ago

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Answered by

alok kumar singh | Contributor-Level 10

4 months ago

81. Total number of balls in the urn = 25

Balls bearing mark ‘X’ = 10

Balls bearing mark ‘Y’ = 15

p = P (ball bearing mark ‘X’) =10/25 = 2/5

q = P (ball bearing mark ‘Y’) =15/25 = 3/5

Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials.

Let Z be the random variable that represents the number of balls with ‘Y’ mark on them in the trials.

Clearly, Z has a binomial distribution with n = 6 and p =2/5.

$∴ P = (Z = z) = {nC}_{z}$
$p^{n - z} q^{z}$

P (all will bear ‘X’ mark) $= P (Z = 0) = {6C}_{0}$
${(\frac{2}{5})}^{6} = {(\frac{2}{5})}^{6}$

P (not more than 2 bear ‘Y’ mark) $= P (Z \leq 2)$

$= P (Z = 0) + P (Z = 1) + P (Z = 2) = {6C}_{0}$
${(p)}^{6} {(q)}^{0} +6 C_{1}$
${(p)}^{5} {(q)}^{1} + {6C}_{2}$
$\begin{array}{l} {(p)}^{4} {(q)}^{2} \\ = {(\frac{2}{5})}^{6} + 6 {(\frac{2}{5})}^{5} (\frac{3}{5}) + 15 {(\frac{2}{5})}^{4} {(\frac{3}{5})}^{2} \\ = {(\frac{2}{5})}^{4} [{(\frac{2}{5})}^{2} + 6 (\frac{2}{5}) (\frac{3}{5}) + 15 {(\frac{3}{5})}^{2}] \\ = {(\frac{2}{5})}^{4} [\frac{4}{25} + \frac{36}{25} + \frac{135}{25}] \\ = {(\frac{2}{5})}^{4} [\frac{175}{25}] \\ = 7 {(\frac{2}{5})}^{4} \end{array}$

P (at least one ball bears

...more

81. Total number of balls in the urn = 25

Balls bearing mark ‘X’ = 10

Balls bearing mark ‘Y’ = 15

p = P (ball bearing mark ‘X’) =10/25 = 2/5

q = P (ball bearing mark ‘Y’) =15/25 = 3/5

Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials.

Let Z be the random variable that represents the number of balls with ‘Y’ mark on them in the trials.

Clearly, Z has a binomial distribution with n = 6 and p =2/5.

$∴ P = (Z = z) = {nC}_{z}$
$p^{n - z} q^{z}$

P (all will bear ‘X’ mark) $= P (Z = 0) = {6C}_{0}$
${(\frac{2}{5})}^{6} = {(\frac{2}{5})}^{6}$

P (not more than 2 bear ‘Y’ mark) $= P (Z \leq 2)$

P (at least one ball bears ‘Y’ mark) $= P (Z \geq 1) = 1 - P (Z = 0)$

$= 1 - {(\frac{2}{5})}^{6}$

P (equal number of balls with ‘X’ mark and ‘Y’ mark) $= P (Z = 3)$

$= {6C}_{3}$
$\begin{array}{l} {(\frac{2}{5})}^{3} {(\frac{3}{5})}^{3} \\ = \frac{20 \times 8 \times 27}{15625} \\ = \frac{864}{3125} \end{array}$

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