When we deal the problem of probability that involves two or more stages, it is frequently useful to use probability tree diagram. To construct the tree diagram, the probability written on each branch of the tree. The probability of an event for the experiment description by a tree is the product of the probabilities attached to the branches along the path representing the event. A tree diagram is the best way of finding events of a sample space with the help of a diagram.
For example:
Tossing three coins simultaneously, the probability tree diagram can be shown as below.
Second way of tree diagram
Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
To construct the tree diagram, the probability written on each branch of the tree. The probability of an event for the experiment description by a tree is the product of the probabilities attached to the branches along the path representing the event. A tree diagram is the best way of finding events of a sample space with the help of a diagram.
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Let us represent the events getting a blue ball by B and a green ball by G. The probability tree diagram is as follows:
Now,
From probability tree diagram,
P(B/B) = P_{1}(B) × P_{2}(B) = \(\frac 37\) × \(\frac 5{13}\) = \(\frac {15}{91}\)_{Ans}
Let W denote a white ball and B denote a black ball.
The probability tree diagram is shown as follows:
Let Y denote yellow ball and B denote the blue ball.
Let us represent all possibilities by probability tree diagram as follows:
Let R, W and Y denote red sweet, white sweet and yellow sweet respectively.
The probability is shown in the following tree diagram:
Now,
Sample space (S) = {(R, W), (R, Y), (W, R), (W, Y), (Y, R), (Y, W)}_{Ans}
Let G denote a green ball and B denote a black ball. The probability tree diagram may be shown as below:
\begin{align*} {\text{The probability of getting both black balls P(B/B)}} &= P_1(B) × P_2(B)\\ &= \frac 9{16} × \frac 8{15}\\ &= \frac 32 × \frac 15\\ &= \frac 3{10}_{Ans}\\ \end{align*}
We represent all the possibilities by the following tree diagram.
\begin{align*} {\text{The probability occurring head in three times P(HHH)}} &= P_1(H) ×P_2(H)× P_3(H)\\ &= \frac 12 ×\frac 12× \frac 12\\ &= \frac 18_{Ans}\\ \end{align*}
Let K denote the event of getting a king and \(\overline{K}\) denote the event of not getting a king.
All the probabilities are shown in the following tree diagram.
Let R and B denote the red and blue marble respectively. The probability tree diagram is shown below:
\begin{align*} {\text{The probability that both marbles are of the same color P(B/B) + P(R/R)}} &= P_1(B) ×P_2(B) + P_1(R) ×P_2(R)\\ &= \frac 47 ×\frac 12 + \frac 37 ×\frac 13\\ &= \frac 27 + \frac 17\\ &= \frac {2 + 1}7\\ &= \frac 37_{Ans}\\ \end{align*}
Let B and G represent a boy and a girl respectively.
Then, the probability tree diagram is shown as follows:
Here,
\begin{align*} P(BBG) + P(BGB) + P(GBB) &= P_1(B) ×P_2(B) ×P_3(G) + P_1(B) ×P_2(G) ×P_3(B) + P_1(G) ×P_2(B) ×P_3(B)\\ &= \frac 12 ×\frac 12 × \frac 12 + \frac 12 × \frac 12 × \frac 12 + \frac 12 × \frac 12 × \frac 12\\ &= \frac 18 + \frac 18 + \frac 18\\ &= \frac 38_{Ans}\\ \end{align*}
Let R and B denote red and blue ball respectively. The probability tree diagram is shown as below:
A bag contains 4 black balls and 3 white balls. A ball is drawn at random and replaced. After that another ball is drawn. Find the probability that both the balls are black.
(frac {49}{16})
1
(frac {16}{49})
(frac {25}{49})
A bag contains 4 black balls and 5 white balls. A ball is drawn at random and replaced. After that another ball is drawn. Find the probability that both the balls are white.
(frac {25}{81})
(frac {16}{25})
(frac {81}{25})
1
A bag contains 3 red and 5 white balls. A ball is drawn at random and replaced. After that another ball is drawn. Find the probability of both are blue.
1
(frac {64}9)
(frac 9{64})
Two cards are drawn from a well shuffled deck of 52 cards. Find the probability that both cards are king (the first card is replaced).
1
(frac {50}{169})
(frac 1{169})
(frac {169}{1})
Two children were born from a married couple. Find the probability of having no daughter.
(frac 41)
(frac 14)
(frac 12)
1
Two children were born from a married couple. Calculate the probability of both are son.
(frac 14)
(frac 12)
(frac 41)
1
A coin is tossed two times in succession. Calculate the probability that both are tails.
(frac 14)
1
(frac 34)
(frac 41)
There are 7 red and 5 green balls of the same shape and size in a basket. If two balls are drawn randomly in succession without replacement from the basket, find the probability of getting both balls of the same colour.
(frac {66}{31})
1
(frac {31}{66})
(frac {3}{51})
Among 400 students of a school, 150 are girls. For dance competition two students are choosen. Find the probability of at least one girl.
1
(frac {1064}{649})
(frac {639}{1034})
(frac {649}{1064})
There are 25 bulbs in a store of which 6 are defective. If two bulbs are drawn randomly in succession without replacement, show the probability of getting both bulbs of the same quality.
(frac {24}{25})
(frac {31}{50})
(frac {50}{31})
1
A bag contains red, white and green marbles in the ratio of 4 : 5 : 6. If two marbles are drawn in succession without replacement. What is the probability of getting both the red marbles?
1
(frac 12)
(frac {35}2)
(frac 2{35})
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aashish poudel
but it refers to replacement and without
Jan 30, 2017
2 Replies
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