29. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that:
(i) Both balls are red.
(ii) First ball is black and second is red.
(iii) One of them is black and other is red.
29. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that:
(i) Both balls are red.
(ii) First ball is black and second is red.
(iii) One of them is black and other is red.
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1 Answer
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29. Total number of balls
Number of red balls
Number of black balls
i. Probability of getting red ball at first draw
The red ball is replaced after the first draw
Probability of getting red ball in second draw
Probability of getting both the balls red
ii. Probability of getting first black ball
The ball is replaced after first draw
So, the probability of getting second ball as red
Probability of getting first ball black and second ball as red
iii. Probability of getting first ball as red
The ball is replaced after the first draw
Probability of getting second ball as black
Therefore, probabili
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Similar Questions for you
P (2 obtained on even numbered toss) = k (let)
P (2) =
P (
If x = 0, y = 6, 7, 8, 9, 10
If x = 1, y = 7, 8, 9, 10
If x = 2, y = 8, 9, 10
If x = 3, y = 9, 10
If x = 4, y = 10
If x = 5, y = no possible value
Total possible ways = (5 + 4 + 3 + 2 + 1) * 2
= 30
Required probability
P (2W and 2B) = P (2B, 6W) × P (2W and 2B)
+ P (3B, 5W) × P (2W and 2B)
+ P (4B, 4W) × P (2W and 2B)
+ P (5B, 3W) × P (2W and 2B)
+ P (6B, 2W) × P (2W and 2B)
(15 + 30 + 36 + 30 + 15)
Let probability of tail is
⇒ Probability of getting head =
∴ Probability of getting 2 heads and 1 tail
ax2 + bx + c = 0
D = b2 – 4ac
D = 0
b2 – 4ac = 0
b2 = 4ac
(i) AC = 1, b = 2 (1, 2, 1) is one way
(ii) AC = 4, b = 4
(iii) AC = 9, b = 6, a = 3, c = 3 is one way
1 + 3 + 1 = 5 way
Required probability =
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