Answer:
Answer is
[tex]\frac{7}{30}[/tex]
Step-by-step explanation:
Given that there are two urns. First urn contains 3 white balls and 6 yellow balls and the second urn contains 7 white balls and 3 yellow balls.
A ball is drawn from each urn.
Probability for ball from urn I is white = no of white balls/total balls in I urn
[tex]= \frac{3}{9} =\frac{1}{3}[/tex]
Probability for ball from urn II is white = no of white balls/total balls in I urn
[tex]=\frac{7}{10}[/tex]
Since drawing one ball from each urn is independent of the other we find
Prob that both balls are white=
Prob that ball from I urn is white *Prob that ball from II urn is white
=[tex]= \frac{7}{10} *\frac{1}{3}\\=\frac{7}{30}[/tex]
The probability that both balls are white is 7/30
The probability of an event is defined as the possibility of an event occurring against sample space.
Let us tackle the problem.
If the first urn contains 3 white balls and 6 yellow balls , then the probability of picking up a white ball from the first urn is:
[tex]P(W_1) = \frac{3}{3 + 6}[/tex]
[tex]P(W_1) = \frac{3}{9}[/tex]
[tex]\large {\boxed {P(W_1) = \frac{1}{3} } }[/tex]
If the second urn contains 7 white balls and 3 yellow balls , then the probability of picking up a white ball from the second urn is:
[tex]P(W_2) = \frac{7}{3 + 7}[/tex]
[tex]\large {\boxed {P(W_2) = \frac{7}{10} } }[/tex]
From the above results , the probability that both balls are white can be calculated using following formula:
[tex]P( W_1 \bigcap W_2 ) = P(W_1) \times P(W_2)[/tex]
[tex]P( W_1 \bigcap W_2 ) = \frac{1}{3} \times \frac{7}{10}[/tex]
[tex]\large {\boxed {P( W_1 \bigcap W_2 ) = \frac{7}{30} } }[/tex]
The probability that both balls are white is 7/30
Grade: High School
Subject: Mathematics
Chapter: Probability
Keywords: Probability , Sample , Space , Six , Dice , Die , Urn , White , Yellow , Ball , Random
