The question
(a) Out of two bags, bag I contains 3 red and 4 white balls and bag II contains 8 red and 6 white balls. A die is thrown. If it shows a number less than 3 then a ball is drawn at random from bag I, otherwise a ball is drawn at random from bag II. Find the probability that the ball drawn from one of the bags is a red ball.
OR
(b) The probability of simultaneous occurrence of atleast one of the two events X and Y is a. If the probability that exactly one of the events X, Y occurs is b, prove that P ( X ′ ) + P ( Y ′ ) = 2 − 2 a + b P(X') + P(Y') = 2 - 2a + b P ( X ′ ) + P ( Y ′ ) = 2 − 2 a + b
Exam answer Show all explanations
(a)
Let E 1 E_1 E 1 E 2 E_2 E 2 A A A
Why?
The die decides which bag is used, and then a ball is drawn. E 1 E_1 E 1 E 2 E_2 E 2
P ( E 1 ) = 2 6 , P ( E 2 ) = 4 6 P(E_1) = \frac26,\quad P(E_2) = \frac46 P ( E 1 ) = 6 2 , P ( E 2 ) = 6 4 Why?
Numbers less than 3 on a die: 1, 2, so 2 of the 6 faces. The rest, 3, 4, 5, 6, are 4 faces.
P ( A ∣ E 1 ) = 3 7 , P ( A ∣ E 2 ) = 8 14 P(A|E_1) = \frac37,\quad P(A|E_2) = \frac{8}{14} P ( A ∣ E 1 ) = 7 3 , P ( A ∣ E 2 ) = 14 8 Why?
P ( A ∣ E 1 ) P(A|E_1) P ( A ∣ E 1 ) 3 + 4 = 7 3 + 4 = 7 3 + 4 = 7 8 + 6 = 14 8 + 6 = 14 8 + 6 = 14
By the theorem of total probability,
P ( A ) = P ( E 1 ) P ( A ∣ E 1 ) + P ( E 2 ) P ( A ∣ E 2 ) = 2 6 ⋅ 3 7 + 4 6 ⋅ 8 14 = 1 7 + 8 21 = 3 + 8 21 \begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac26\cdot\frac37 + \frac46\cdot\frac{8}{14} \\ &= \frac17 + \frac{8}{21} = \frac{3 + 8}{21} \end{aligned} P ( A ) = P ( E 1 ) P ( A ∣ E 1 ) + P ( E 2 ) P ( A ∣ E 2 ) = 6 2 ⋅ 7 3 + 6 4 ⋅ 14 8 = 7 1 + 21 8 = 21 3 + 8 Why?
A red ball comes either from bag I or from bag II. Each route is "choose the bag, then draw red", so multiply along the route and add the two routes.
The arithmetic:
2 6 ⋅ 3 7 = 2 × 3 6 × 7 = 6 42 = 6 ÷ 6 42 ÷ 6 = 1 7 4 6 ⋅ 8 14 = 4 × 8 6 × 14 = 32 84 = 32 ÷ 4 84 ÷ 4 = 8 21 \begin{aligned} \frac26\cdot\frac37 &= \frac{2 \times 3}{6 \times 7} = \frac{6}{42} \\ &= \frac{6 \div 6}{42 \div 6} = \frac17 \\[4pt] \frac46\cdot\frac{8}{14} &= \frac{4 \times 8}{6 \times 14} = \frac{32}{84} \\ &= \frac{32 \div 4}{84 \div 4} = \frac{8}{21} \end{aligned} 6 2 ⋅ 7 3 6 4 ⋅ 14 8 = 6 × 7 2 × 3 = 42 6 = 42 ÷ 6 6 ÷ 6 = 7 1 = 6 × 14 4 × 8 = 84 32 = 84 ÷ 4 32 ÷ 4 = 21 8 Then 1 7 = 3 21 \frac17 = \frac{3}{21} 7 1 = 21 3
1 7 = 1 × 3 7 × 3 = 3 21 3 21 + 8 21 = 3 + 8 21 = 11 21 \begin{aligned} \frac17 &= \frac{1 \times 3}{7 \times 3} = \frac{3}{21} \\ \frac{3}{21} + \frac{8}{21} &= \frac{3 + 8}{21} = \frac{11}{21} \end{aligned} 7 1 21 3 + 21 8 = 7 × 3 1 × 3 = 21 3 = 21 3 + 8 = 21 11 ∴ \therefore ∴ 11 21 \frac{11}{21} 21 11
(b)
Given P ( X ∪ Y ) = a P(X \cup Y) = a P ( X ∪ Y ) = a
Why?
"At least one of X and Y occurs" is the event X ∪ Y X \cup Y X ∪ Y
P(exactly one of X, Y occurs) = P ( X ∪ Y ) − P ( X ∩ Y ) = P(X \cup Y) - P(X \cap Y) = P ( X ∪ Y ) − P ( X ∩ Y )
Why?
X ∪ Y X \cup Y X ∪ Y X ∩ Y X \cap Y X ∩ Y
a − P ( X ∩ Y ) = b ⇒ P ( X ∩ Y ) = a − b \begin{aligned} a - P(X \cap Y) &= b \\ \Rightarrow P(X \cap Y) &= a - b \end{aligned} a − P ( X ∩ Y ) ⇒ P ( X ∩ Y ) = b = a − b Why?
Add P ( X ∩ Y ) P(X \cap Y) P ( X ∩ Y ) b b b
a = b + P ( X ∩ Y ) a − b = P ( X ∩ Y ) \begin{aligned} a &= b + P(X \cap Y) \\ a - b &= P(X \cap Y) \end{aligned} a a − b = b + P ( X ∩ Y ) = P ( X ∩ Y ) Using P ( X ) + P ( Y ) P(X) + P(Y) P ( X ) + P ( Y ) = P ( X ∪ Y ) + P ( X ∩ Y ) = P(X \cup Y) + P(X \cap Y) = P ( X ∪ Y ) + P ( X ∩ Y )
Why?
This is the addition theorem rearranged:
P ( X ∪ Y ) = P ( X ) + P ( Y ) − P ( X ∩ Y ) \begin{aligned} P(X \cup Y) &= P(X) + P(Y) \\ &\quad - P(X \cap Y) \end{aligned} P ( X ∪ Y ) = P ( X ) + P ( Y ) − P ( X ∩ Y ) Move P ( X ∩ Y ) P(X \cap Y) P ( X ∩ Y )
P ( X ′ ) + P ( Y ′ ) = [ 1 − P ( X ) ] + [ 1 − P ( Y ) ] = 2 − [ P ( X ) + P ( Y ) ] = 2 − [ P ( X ∪ Y ) + P ( X ∩ Y ) ] = 2 − [ a + ( a − b ) ] \begin{aligned} &P(X') + P(Y') \\ &= [1 - P(X)] + [1 - P(Y)] \\ &= 2 - [P(X) + P(Y)] \\ &= 2 - [P(X \cup Y) + P(X \cap Y)] \\ &= 2 - [a + (a - b)] \end{aligned} P ( X ′ ) + P ( Y ′ ) = [ 1 − P ( X )] + [ 1 − P ( Y )] = 2 − [ P ( X ) + P ( Y )] = 2 − [ P ( X ∪ Y ) + P ( X ∩ Y )] = 2 − [ a + ( a − b )] Why?
The complement rule: P ( X ′ ) = 1 − P ( X ) P(X') = 1 - P(X) P ( X ′ ) = 1 − P ( X ) Y Y Y
Then a + ( a − b ) = 2 a − b a + (a - b) = 2a - b a + ( a − b ) = 2 a − b 2 − ( 2 a − b ) 2 - (2a - b) 2 − ( 2 a − b ) − ( − b ) = + b -(-b) = +b − ( − b ) = + b
2 − [ a + ( a − b ) ] = 2 − [ 2 a − b ] = 2 − 2 a + b \begin{aligned} &2 - [a + (a - b)] \\ &= 2 - [2a - b] \\ &= 2 - 2a + b \end{aligned} 2 − [ a + ( a − b )] = 2 − [ 2 a − b ] = 2 − 2 a + b ∴ P ( X ′ ) + P ( Y ′ ) = 2 − 2 a + b \therefore\ P(X') + P(Y') = 2 - 2a + b ∴ P ( X ′ ) + P ( Y ′ ) = 2 − 2 a + b
Where this comes from Chapter 13, Probability : Total probability; probability of events.
Question type: (a) Two routes to one result
How to spot it: click to see (a) Type 7. Two routes to one result: total probability
Spot it: the result can happen in two (or more) ways, such as a ball from bag I or from bag II, and you want its overall chance.
Here: The red ball can come from bag I (the die shows 1 or 2, chance 2 6 \frac26 6 2 4 6 \frac46 6 4 3 7 \frac37 7 3 8 14 \frac{8}{14} 14 8 11 21 \frac{11}{21} 21 11
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