JEE MAIN - Mathematics (2009 - No. 7)
One ticket is selected at random from $$50$$ tickets numbered $$00, 01, 02, ...., 49.$$ Then the probability that the sum of the digits on the selected ticket is $$8$$, given that the product of these digits is zer, equals :
$${1 \over 7}$$
$${5 \over 14}$$
$${1 \over 50}$$
$${1 \over 14}$$
Explanation
Sample space = {00, 01, 02, 03, ..........49} = 50 tickets
n(S) = 50
n(Sum = 8) = { 08, 17, 26, 35, 44 } = 5
n(Product = 0) = { 00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40 } = 14
$$\therefore$$ Probability when product is 0 = P(Product = 0) = $${14 \over {50}}$$
n(Sum = 8 $$ \cap $$ Product = 0) = { 08 } = 1
$$\therefore$$ Probability when sum is 8 and product is 0 = P(Sum = 8 $$ \cap $$ Product = 0) = $${1 \over {50}}$$
Required probability,
$$P\left( {{{Sum = 8} \over {Product = 0}}} \right)$$
= $${{P\left( {Sum = 8 \cap Product = 0} \right)} \over {P\left( {Product = 0} \right)}}$$
= $${{{1 \over {50}}} \over {{{14} \over {50}}}}$$
=$${1 \over {14}}$$
$$\therefore$$ Option (D) is correct.
n(S) = 50
n(Sum = 8) = { 08, 17, 26, 35, 44 } = 5
n(Product = 0) = { 00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40 } = 14
$$\therefore$$ Probability when product is 0 = P(Product = 0) = $${14 \over {50}}$$
n(Sum = 8 $$ \cap $$ Product = 0) = { 08 } = 1
$$\therefore$$ Probability when sum is 8 and product is 0 = P(Sum = 8 $$ \cap $$ Product = 0) = $${1 \over {50}}$$
Required probability,
$$P\left( {{{Sum = 8} \over {Product = 0}}} \right)$$
= $${{P\left( {Sum = 8 \cap Product = 0} \right)} \over {P\left( {Product = 0} \right)}}$$
= $${{{1 \over {50}}} \over {{{14} \over {50}}}}$$
=$${1 \over {14}}$$
$$\therefore$$ Option (D) is correct.
Comments (0)
