What is the probability of getting all three same outcomes when an unbiased coin is tossed thrice 1 point?

A coin is tossed three times. Find the probability of the following eventsA: getting at least two heads,B: getting exactly two heads,C: getting at most one head,D: getting more heads than tails.

Answer

Hint:
Collect all the possible combinations for sample space. From that we get the total number of possible outcomes. Then we get the probability by using formula.

Complete step by step solution:
The coin is tossed three times the possible outcome number is $2\times 2\times 2=8$. The number two is the possibility of heads and tails when it is tossed three times we need to multiply by three times. Now each individual coin is equally likely to come up heads or tails then we have possible combinations are HHH, HHT, HTT, HTH, TTT, TTH, THH and THT. These are the sample space, sample space is nothing but a set of all possible outcomes of the experiment or results of that experiment. We denoted sample space in set notation, it is represented by the symbol “S''.
So we can write sample space in this format,
S= {HHH, HHT, HTT, HTH, TTT, TTH, THH, THT}
To find the probability we need to divide the favorable outcomes by the number of outcomes. A favorable outcome means that the number of events we want from the experiment.
$\Rightarrow $ Probability of A = favorable outcomes ÷number of outcomes
1) A: getting at least two heads,
Here we need the probability of getting at least two heads, at least two heads means getting two heads and more than that. To determine the results we need to check from the sample space which gives at least two heads.
HHH-yes, HHT-yes, HTT-no, HTH-yes, TTT-no, TTH-no, THH-yes, THT-no.
$\Rightarrow $ Here the favorable outcomes are HHT, HTH, THH and HHH.
$\Rightarrow $ Number of favorable outcomes is $4$
$\Rightarrow $ Number of outcomes here is $8$
$\Rightarrow P(A)=\dfrac{4}{8}$
$\Rightarrow P(A)=\dfrac{1}{2}$
Probability of getting at least two heads is $\dfrac{1}{2}$
2) B: getting exactly two heads,
Here we need to find the probability of getting exactly two heads means we have to get absolutely two heads. To determine the results we have to check from the sample space,
HHH-no, HHT-yes, HTT-no, HTH-yes, TTT-no, TTH-no, THH-yes, THT-no.
$\Rightarrow $ Here the favorable outcomes are HHT, HTH and THH.
$\Rightarrow $ Number of favorable outcomes is $3$
So here we have the probability as,
$\Rightarrow P(B)=\dfrac{3}{8}$
Probability of getting exactly two heads is $\dfrac{3}{8}$
3) C: getting at most one head,
Here we need to find the probability of getting at most one head.
At most one head means that only one of the three coins show up ahead or no head. To determine the results we have to check from the sample space,
HHH-no, HHT-no, HTT-yes, HTH-no, TTT-yes, TTH-yes, THH-no, THT-yes.
$\Rightarrow $ Here the favorable outcomes are HTT, TTT, TTH and THT.
$\Rightarrow $ Number of favorable outcomes is $4$
So apply the formula we get,
$\Rightarrow P(C)=\dfrac{4}{8}$
We can also write,
$\Rightarrow P(C)=\dfrac{1}{2}$
Probability of getting at most one head is $\dfrac{1}{2}$
4) D: getting more heads than tails,
Here we need to find the probability of getting more heads than tails.
This means the heads come up is more than tails come up. To determine the results we have to check from the sample space,
HHH-yes, HHT-yes, HTT-no, HTH-yes, TTT-no, TTH-no, THH-yes, THT-no.
$\Rightarrow $ Here the favorable outcomes are HHH, HHT, HTH and THH.
$\Rightarrow $ Number of favorable outcomes is $4$
So we get,
$\Rightarrow P(D)=\dfrac{4}{8}$
Then we have,
$\Rightarrow P(D)=\dfrac{1}{2}$
Probability of getting more heads than tails is $\dfrac{1}{2}$

Note:
A random experiment is an experiment in which
1) The set of all possible outcomes are known.
2) Exact outcome is not known.

This set of Probability and Statistics Assessment Questions and Answers focuses on “Set Theory of Probability – 2”.

1. If P(B⁄A) = p(b), then P(A ∩ B) = ____________
a) p(b)
b) p(a)
c) p(b).p(a)
d) p(a) + p(b)
View Answer

Answer: c
Explanation: P(B/A) = p(b) implies A and B are independent events
Therefore, P(A ∩ B) = p(a).p(b).

2. Two unbiased coins are tossed. What is the probability of getting at most one head?
a) 1⁄2
b) 1⁄3
c) 1⁄6
d) 3⁄4
View Answer

Answer: d
Explanation: Total outcomes = (HH, HT, TH, TT)
Favorable outcomes = (TT, HT, TH)
At most one head refers to maximum one head,
Therefore, probability = 3⁄4.

3. If A and B are two events such that p(a) > 0 and p(b) is not a sure event, then P(A/B) = ?
a) 1 – P(A /B)
b) \(P\frac{(\bar{A})}{(\bar{B})}\)
c) Not Defined
d) \(\frac{1-P(A \cup B)}{P(\bar{B}}\)
View Answer

Answer: d
Explanation: From definition of conditional probability we have
\(P(\bar{A}/\bar{B})=\frac{\bar{A}\cap\bar{B}}{P(\bar{B})}\)
Using De Morgan’s Law
=\(\frac{P(\bar{A \cup B})}{(\bar{B}}\)
=\(\frac{1-P(A \cup B)}{P(\bar{B}}\)

4. If A and B are two events, then the probability of exactly one of them occurs is given by ____________
a) P(A ∩ B) + P(A ∩ B)
b) P(A) + P(B) – 2P(A) P(B)
c) P(A) + P(B) – 2P(A) P(B)
d) P(A) + P(B) – P(A ∩ B)
View Answer

Answer: a
Explanation: The set corresponding to the required outcome is
(A ∩ B) ∪ (A ∩ B)
Hence the required probability is
P(P(A ∩ B) ∪ (A ∩ B)) = (A ∩ B) + P(A ∩ B).

5. The probability that at least one of the events M and N occur is 0.6. If M and N have probability of occurring together as 0.2, then P(M) + P(N) is?
a) 0.4
b) 1.2
c) 0.8
d) Indeterminate
View Answer

Answer: b
Explanation: Given : P(M ∪ N) = 0.6, P(M ∩ N) = 0.2
P(M ∪ N) + P(M ∩ N) = P(M) + P(N)
2 – (P(M ∪ N) + P(M ∩ N)) = 2 – (P(M) + P(N))
= (1 – P(M)) + (1 – P(N))
2 – (0.6 + 0.2) = P(M) + P(N)
P(M) + P(N) = 2 – 0.8
= 1.2

6. A jar contains ‘y’ blue colored balls and ‘x’ red colored balls. Two balls are pulled from the jar without replacing. What is the probability that the first ball Is blue and second one is red?
a) \(\frac{xy-y}{x^2+y^2+2xy-(x+y)}\)
b) \(\frac{xy}{x^2+y^2+2xy-(x+y)}\)
c) \(\frac{y^2-y}{x^2+y^2+2xy-(x+y)}\)
d) \(\frac{xy-y}{x^2+y^2+2xy-(x-y)}\)
View Answer

Answer: b
Explanation: Number of blue balls = y
Number of Red balls = x
Total number of balls = x + y
Probability of Blue ball first = y / x + y
No. of balls remaining after removing one = x + y – 1
Probability of pulling secondball as Red=\(\frac{x}{x+y-1}\)
Required porbability=\(\frac{y}{(x+y)}\frac{x}{(x+y-1)}=\frac{xy}{x^2+y^2+2xy-(x+y)}\)

7. A survey determines that in a locality, 33% go to work by Bike, 42% go by Car, and 12% use both. The probability that a random person selected uses neither of them is?
a) 0.29
b) 0.37
c) 0.61
d) 0.75
View Answer

Answer: b
Explanation: Given: p(b) = 0.33, P(c) = 0.42
P(B ∩ C) = 0.12
P(B ∩ C) = ?
P(B ∩ C) = 1 – P(B ∪ C)
= 1 – p(b) – P(c) + P(B ∩ C)
= 1 – 0.22 – 0.42 + 0.12
= 0.37.

8. A coin is biased so that its chances of landing Head is 2⁄3. If the coin is flipped 3 times, the probability that the first 2 flips are heads and the 3rd flip is a tail is?
a) 4⁄27
b) 8⁄27
c) 4⁄9
d) 2⁄9
View Answer

Answer: a
Explanation: Required probability = 2⁄3 x 2⁄3 x 1⁄3 = 4⁄27.

9. Husband and wife apply for two vacant spots in a company. If the probability of wife getting selected and husband getting selected are 3/7 and 2/3 respectively, what is the probability that neither of them will be selected?
a) 2⁄7
b) 5⁄7
c) 4⁄21
d) 17⁄21
View Answer

Answer: c
Explanation:Let H be the event of husband getting selected
W be the event of wife getting selected
Then, the event of neither of them getting selected is = (H ∩ W)
P (H ∩ W) = P (H) x P (W)
= (1 – P (H)) x (1 – P (W))
= (1 – 2⁄3) x (1 – 3⁄7)
= 4⁄21.

10. For two events A and B, if P (B) = 0.5 and P (A ∪ B) = 0.5, then P (A|B) = ?
a) 0.5
b) 0
c) 0.25
d) 1
View Answer

Answer: d
Explanation: We know that,
P (A│B) = P(A ∩ B)/P(B)
= P((A ∪ B)/P(B))
= (1 – P(A ∪ B)) /P(B)
= (1 – 0.5)/0.5
= 1.

11. A fair coin is tossed thrice, what is the probability of getting all 3 same outcomes?
a) 3⁄4
b) 1⁄4
c) 1⁄2
d) 1⁄6
View Answer

Answer: b
Explanation:S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
All 3 same outcomes mean either all head or all tail
Total outcomes = 8
Favourable outcomes = {HHH, TTT} = 2
∴ required probability = 2⁄8 = 1⁄4.

12. A bag contains 5 red and 3 yellow balls. Two balls are picked at random. What is the probability that both are of the same colour?
a) \(\frac{C_2^5}{C_2^8}+\frac{C_2^3}{C_2^8}\)
b) \(\frac{C_2^5 * C_2^3}{C_2^8}\)
c) \(\frac{C_1^5 * C_1^3}{C_2^8}\)
d) 0.5
View Answer

Answer: a
Explanation:Total no.of balls = 5R+3Y = 8
No.of ways in which 2 balls can be picked = 8C2
Probability of picking both balls as red = 5C2 /8C2
Probability of picking both balls as yellow = 3C2 /8C2
∴ required probability \(\frac{C_2^5}{C_2^8}+\frac{C_2^3}{C_2^8}\).

Sanfoundry Global Education & Learning Series – Probability and Statistics.

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