Question

Three letters are chosen at random from the word HEART and arranged in a row. Find the probability that:
a) the letter H is chosen
b) both vowels are chosen

Answer 1

The probability of H being one of the three letters chosen = \(\Large {3 \over 5}\).
The probability of H being selected in one of the three = \(\Large {1 \over 3}\)

So, the answer to (a) will be \({\large{3 \over 5}} \times {\large {1 \over 3}}\).

Answer 2

The probability that a vowel is chosen on the first pick is \(\Large {2 \over 5}\), and in the second pick = \(\Large {1 \over 4}\).

Answer 3

So, the answer to (b) is \(\Large {2 \over 5} \times {1 \over 4}\)

Answer 4

wait how is the probability of H being one of the three letters chosen=3/5? isn't it 1/5?

Answer 5

You are right. Oopsie.

Answer 6

nope, I am not actually, I am just trying to understand how you got it.

Answer 7

and for part b the answer is 0.3 not 0.1 which you got.

Answer 8

The probabilities of two independent events are multiplied.

Answer 9

What? I can't be trusted any more. Sorry!

Answer 10

\(5C3\)

Answer 11

Now H is chosen so fix H first..

And the other two words can be select from 5 letters in:

\[\large ^1C_1 \times ^5C_2 = ??\]

Answer 12

I think it is also 10..

Answer 13

My mind is not working I think..

Answer 14

lol then I must not have one at all.

Answer 15

See you have chosen 3 and put them in a row..

So they can be one out of the 5..

So parth is right to A part I guess..

Answer 16

yes. he did get the correct answer which is 0.2 but so
"The probability of three letters chosen from the 5 letter word Heart = 3/5.
The probability of H being selected in one of the three = 1/3
and we just times them together 3/5 x 1/3 = 3/15 = 1/5 = 0.2

Answer 17

Yes..

Answer 18

Now from 5 we can choose 2 vowels in how many ways ???

Answer 19

1/5 x 1/5 = 1/25?

Answer 20

So:

from 3 we have to choose 2 vowels now yes no??

Answer 21

yes

Answer 22

So 1/9 I guess..

Answer 23

@Ganpat help here in B part..

Answer 24

probability for choosing 3 letters from 5 = 5C3

such that, both vowels r chosen = 2C2... as there r only two vowels in equation
and remaining letter would be 3C1..

So the probability = ( 3C1 * 2C2 ) / 5C3 ..

does this makes sense ??

Answer 25

So the probability = ( 3C1 * 2C2 ) / 5C3 = ( 3 * 1) / (5 * 4 * 3)/(3 * 2) = 3/10

3/10, is probability for part b... recheck my calculations..

Answer 26

i think one way to do first part is
the possibilities are
HNN
NHN
NNH

where N is not a H

probability is
1/5 for first
4/5 * 1/4 for second
and 4/5*3/4 * 1/3 for third

add these up to get answer

Answer 27

= 1/5 + 4/20 + 12/60 = 3/5
- i'm not absolutely sure thats valid

Answer 28

does that make sense to you jays? probability is not my strong point.

Answer 29

hmm not really and just a heads up that 3/5 is not the answer for part a.

Answer 30

I guess I understand @Ganpat's method even though it's a bit confusing and I might forget how to apply how he did it to other similar questions.

Answer 31

do you have the answers?

Answer 32

correct answers?

Answer 33

actually ur right, my bad.

Answer 34

part a) 0.6
part b) 0.3

Answer 35

ok then ganpat was right about part b

Answer 36

Three letters are chosen at random from the word HEART and arranged in a row. Find the probability that:
a) the letter H is chosen

We assume no replacement. Order does not matter (so combinations)
there are 5C3 combinations in total.
to find the number with an H, we can write down
H _ _ where we have fixed the first choice to H. there are 4C2 ways to fill the 2 empty slots. Probability is
\[\frac{\left(\begin{matrix} 4\\ 2\end{matrix}\right)}{\left(\begin{matrix}5 \\ 3\end{matrix}\right)}\] or 0.6

Answer 37

Three letters are chosen at random from the word HEART and arranged in a row. Find the probability that:
b) both vowels are chosen

As above, the total number of patterns is 5C3= 10
if both vowels are chosen, we have
A E _ (order does not matter)
we can fill the last slot 3C1 different ways (5-2 leaves 3 letters to choose from)
we get 3/10 = 0.3

Answer 38

If we want to be pedantic, there is 1C1 way to choose A, 1C1 way to choose E, and 3C1 to pick the last letter, so the numerator is
1C1 * 1C1 * 3C1