Question 71536: 1. A coin is tossed thrice. What is the probability that at least 2 heads occur?
2. A pair of dice is rolled. What is the probability that the sum is equal to:
a. 5
b. 10
3. If a card is drawn at random from an ordinary deck of 52 cards, find the probability that it is:
a. Diamond
b. A red card
c. A heart or a spade
d. Not an ace
e. A black card or a king
4. The letters of the word HONESTY are written in slips of paper, and are placed in a box. A slip of paper is chosen at random. What is the probability that:
a. The letter is a vowel
b. The letter is a consonant
c. The letter is H
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
1. A coin is tossed thrice. What is the probability that at least 2 heads occur?
Since it's heads or tails, it's binomial.
P(at least 2 heads) = P(2 heads)+P(3 heads) = [3C2(1/2)2(1/2)=3(1/8)=3/8] +
[3C3(1/2)3(1/2)^0] = 1*1/8*1= 1/8 = 4/8= 1/2
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2. A pair of dice is rolled. What is the probability that the sum is equal to:
a. 5
The sample space has (1,4),(2,3),(3,2),(4,1)
P(5)= 4/36 = 1/9
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b. 10
The sample space has (4,6),(5,5),(6,4)
P(10)=3/36 = 1/12
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3. If a card is drawn at random from an ordinary deck of 52 cards, find the probability that it is:
a. Diamond: 13/52 = 1/4
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b. A red card: 26/52 = 1/2
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c. A heart or a spade: 13/52 + 13/52 = 26/52 = 1/2
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d. Not an ace; 1-P(ace) = 1 - 4/52 = 48/52
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e. A black card or a king: P(black)+P(king)-P(black and king)
=26/52 + 4/52 -2/52 = 28/52
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4. The letters of the word HONESTY are written in slips of paper, and are placed in a box. A slip of paper is chosen at random. What is the probability that:
a. The letter is a vowel: 2/7
b. The letter is a consonant: 5/7
c. The letter is H: 1/7
==========================
Cheers,
Stan H.
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting exactly two heads.
A = {HHT, HTH, THH}
n(A) = 3
P(A) = `("n"("A"))/("n"("S")) = 3/8`
Let B be the event of getting atleast one tail
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
P(B) = `("n"("B"))/("n"("S")) = 7/8`
Let C be the event of getting consecutively
C = {HHH, HHT, THH}
n(C) = 3
P(C) = `("n"("C"))/("n"("S")) = 3/8`
A ∩ B = {HHT, HTH, THH}
n(A ∩ B) = 3
p(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 3/8`
B ∩ C = {HHT, THH}
n(B ∩ C) = 2
P(B ∩ C) = `("n"("B" ∩ "C"))/("n"("S")) = 2/8`
A ∩ C = {HHT, THH}
n(A ∩ C) = 2
P(A ∩ C) = `("n"("A" ∩ "C"))/("n"("S")) = 2/8`
(A ∩ B ∩ C) = {HHT, THH}
n(A ∩ B ∩ C) = 2
P(A ∩ B ∩ C) = `("n"("A" ∩ "B" ∩ "C"))/("n"("S")) = 2/8`
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C)
= `3/8 + 7/8 + 3/8 - 3/8 - 2/8 - 2/8 + 2/8`
= `3/8 + 7/8 - 2/8`
= `(10 - 2)/8`
= `8/8`
= 1
The probability is 1.
The ratio of successful events A = 4 to the total number of possible combinations of a sample space S = 8 is the probability of 2 heads in 3 coin tosses. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 2 heads, if a coin is tossed three times or 3 coins tossed together. Users may refer this tree diagram to learn how to find all the possible combinations of sample space for flipping a coin one, two, three or four times.
Solution
Step by step workout
step 1 Find the total possible events of sample space S
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
S = 8
step 2 Find the expected or successful events A
A = {HHH, HHT, HTH, THH}
A = 4
step 3 Find the probability
P(A) = Successful Events/Total Events of Sample Space
= 4/8
= 0.5
P(A) = 0.5
0.5 is the probability of getting 2 Heads in 3 tosses.
Probability is the measurement of chances – the likelihood that an event will occur. If the probability of an event is high, it is more likely that the event will happen. It is measured between 0 and 1, inclusive. So if an event is unlikely to occur, its probability is 0. And 1 indicates the certainty for the occurrence.
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of outcomes and the total number of outcomes.