What is the probability of getting the letter?

Solution:

Given, the cards bearing letters of the word “MATHEMATICS” are placed in a bag.

A card is taken out from the bag without looking into the bag at random.

(a) We have to find the number of possible outcomes when a letter is taken out of the bag at random.

Number of letters in the given word is 11.

So, there are 11 cards each representing the letters of the given word.

Possible outcomes = M, A, T, H, E, M, A, T, I, C, S

Number of possible outcomes = 11

Therefore, the number of possible outcomes is 11.

(b) Possible outcomes = M, A, T, H, E, M, A, T, I, C, S

Number of possible outcomes = 11

Probability = Number of outcomes favourable to the event/Total number of outcomes in the experiment

(i) We have to find the probability of getting M.

Favourable outcomes = M, M

Number of favourable outcomes = 2

Probability of getting M = 2/11

(ii) We have to find the probability of getting any vowel.

Favourable outcomes = A, E, A, I

Number of favourable outcomes = 4

Probability of getting any vowel = 4/11

(iii) We have to find the probability of getting any consonant.

Favourable outcomes = M, T, H, M, T, C, S

Number of favourable outcomes = 7

Probability of getting any consonant = 7/11

(iv) We have to find the probability of getting x.

Favourable outcomes = 0

Number of favourable outcomes = 0

Probability of getting any consonant = 0/11

✦ Try This: The cards bearing letters of the word “TRANSVERSAL” are placed in a bag. A card is taken out from the bag without looking into the bag (at random). How many outcomes are possible when a letter is taken out of the bag at random

☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 3

NCERT Exemplar Class 7 Maths Chapter 3 Sample Problem 11

The cards bearing letters of the word “MATHEMATICS” are placed in a bag. A card is taken out from the bag without looking into the bag (at random). (a) How many outcomes are possible when a letter is taken out of the bag at random? (b) What is the probability of getting (i) M? (ii) Any vowel? (iii) Any consonant? (iv) X

Summary:

The cards bearing letters of the word “MATHEMATICS” are placed in a bag. A card is taken out from the bag without looking into the bag (at random), (a) The number of possible outcomes when a letter is taken out of the bag at random is 11, (b) The probability of getting (i) M is 2/11, (ii) Any vowel is 4/11, (iii) Any consonant is 7/11, (iv) X is 0/11

☛ Related Questions:

• If the mean of 26, 28, 25, x, 24 is 27, find the value of x
• The mean of 10 observations was calculated as 40. It was detected on rechecking that the value of 45 . . . .
• The median of observations 11, 12, 14, 18, x + 2, 20, 22, 25, 61 arranged in ascending order is 21. . . . .

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If every letter in the alphabet is available and repetition is allowed, our sample space consists of all "four letter strings" which includes: aaaa, aaab, aaac, ..., abaa, abab,... zzzz, for a total of $26^{4}$ different available "words."

To calculate the probability that the first letter in the word is in fact a "b", you can count how many four letter strings begin with a "b" and divide by the total number of possible strings. As such $Pr(\text{first is a "b"}) = \frac{26^3}{26^4} = \frac{1}{26}$. (equivalently, you can notice that the remaining spaces in the word don't matter)

Similarly the other spaces will have probability $\frac{1}{26}$ to also be filled with their respective desired letters.

The probability of ending with the word "bake" in this scenario is, as you correctly guessed $(\frac{1}{26})^4$, as each "word" in our sample space is equally likely, and only one occurrence of the word "bake".

If letters are not allowed to be repeated, but all letters are available for use, our sample space then is: abcd, abce, abcf,... bacd, bace,... zyxw for a total of $26\cdot 25\cdot 24\cdot 23$ different "words" available. The probability of each letter individually appearing in their correct spots remains $\frac{1}{26}$, however, the sample space size is smaller, only $26\cdot 25\cdot 24\cdot 23$. As such the probability is no longer $(\frac{1}{26})^4$ but is instead $\frac{1}{26\cdot 25\cdot 24\cdot 23}$. This can be used as proof then that the events "first slot is b", "second slot is a", etc... are not independent from one another.

If the only letters available to use are "a,b,e,k" with letters being allowed to be repeated then our sample space consists of words: aaaa, aaab, aaae, aaak, aaba, aabb, aabe,..., kkke, kkkk, for a total of $4^4$ different such words.

Similarly to above we calculate the probability for each individual letter to appear in its respective place to be $\frac{1}{4}$, and the probability of ending with the word "bake" as $\frac{1}{4^4}$

If our available letters are again "a,b,e,k" and letters are not allowed to be repeated, then our sample space consists of the words: abek, abke, aebk, aekb, akbe, akeb,..., keba, for a total of $4\cdot 3\cdot 2\cdot 1$ different words available.

Again, similarly to before, we calculate the probability of a letter appearing in its desired location to be $\frac{1}{4}$, and the probability of ending with the word "bake" as $\frac{1}{4\cdot 3\cdot 2\cdot 1} = \frac{1}{24}$, which again can be used as proof that the events are not independent.