Question

A group of three-year-olds are presented with three photographs showing a dog, a cat, and a horse as part of a cognitive test. Then they are given the words dog, cat and horse to match with the photographs.

17. Assuming that the children have not yet learned to associate the words with the images, what is the probability of obtaining a completely incorrect match?
A. 0
b. 0.16
c. 0.33
d. 0.50
E. None of the above

18. Assuming that children have not yet learned to associate words with images, what is the probability that they will correctly match a single word and photo pair?
A. 0
b. 0.16
c. 0.33
d. 0.50
E. None of the above

19. Assuming that the children have not yet learned to associate the words with the images, what is the probability that exactly two pairs of photo and word will match?
A. 0
b. 0.16
c. 0.33
d. 0.50
E. None of the above

20. Assuming that the children have not yet learned to associate the words with the images, what is the probability that they will match the three photos and three words correctly?
A. 0
b. 0.16
c. 0.33
d. 0.50
E. None of the above

21. If X: number of correct matches then E (X) is
A. 0
b. one
c. two
d. 3
E. None of the above

22. If X: number of correct matches then its variance is
A. 0
b. one
c. two
d. 3
E. None of the above

23. If X: number of correct pairings then its standard deviation is
A. 0
b. one
c. two
d. 3
E. None of the above

24. If you take a group of 30 children who have not yet learned to associate the words with the images, approximately how many children would you expect to correctly match a single word and photo pair?
A. 11
b. 12
c. 13
d. 14
E. None of the above

Answer 1

Solution-

According to question, a group of 3-year-olds children are presented with 3 photographs showing a dog, a cat, and a horse as part of a cognitive test.

Then they are given the words dog, cat and horse to match with the photographs.

Now, let the events of selecting photographs of dog , cat and horse are denoted by D, C and H respectively and

let the events of selecting the words dog, cat and horse are denoted by d, c and h respectively.

Now, if a photo and a word is randomly sekected then,

Possible outcomes for first selection are 3×3 = 9.

{Dd, Dc, Dh, Cd, Cc, Ch, Hd, Hc, Hh}

So, there 3 correct pairs and 6 incorrect pairs

So, probability of getting correct pair in I^st selection is

P(Ist pair correct) = 3/6 = 1/3 ....(1)

And probability of selecting incorrect pair in I^st is

P(Ist pair incorrect ) = 1 -1/3 = 2/3 ....(2)

Now,

If Ist selected pair is correct , then 2×2 =4 possible pairs are there. Out of which 2 pairs are correct and two are incorrect.

So, P(IInd pair is correct/Ist pair correct)= 2/4 = 1/2 ...(3)

And P(IInd pair is incorrect/Ist pair correct)= 2/4 = 1/2 ...(4)

If Ist selected pair is incorrect, then 2×2 = 4 pairs are remaining .Out of these 4 pairs only 1 is correct.

So, P(IInd pair correct/Ist pair incorrect ) = 1/4 ....(5)

And P(IInd pair incorrect/Ist pair is incorrect) = 1- 1/4 = 3/4 ....(6)

Now,

If first and second pair are correct then 3rd pair will also definitely correct.

So, P(IIIrd pair correct/Ist and IInd correct)=1 ....(7)

And P(IIrd pair incorrect/Ist and IInd correct)=0 ....(8)

And if any of the previous two pairs is/are incorrect then 3 pair will be definitely incorrect.

So, P(IIIId pair incorrect/Ist or IInd incorrect)=1...(9)

And P(IIIrd pair correct/Ist or IInd incorrect)=0 ...(10)

Now, let us use above 10 Probabilities, let us answer the following problems-

(17)

Since we are assuming that the children have not yet learned to associate the words with the images, So the probability of obtaining a completely incorrect match

=P(Ist incorrect)×P(Ist incorrect/IInd incorrect)×P(IIrd incorrect/Ist and IInd incorrect)

=(2/3)×(3/4)×(1) = 2/4 = 1/2 = 0.5 (option-d)

(18)

Since we are assuming that children have not yet learned to associate words with images. So, the probability that they will correctly match a single word and photo pair is

=P(only Ist correct) + P( only IInd correct) + P(only IIIrd correct)

=(1/3)×(1/2)×(1) + (2/3)×(1/4)×(1) + (2/3)×(3/4)×(0)

= 1/6 + 1/6 +0

= 2/6

=1/3 = 0.33 (option - c)

(19)

If two pairs are correctly match then third will also definitely match.So,

Assuming that the children have not yet learned to associate the words with the images, the probability that exactly two pairs of photo and word will match is

Same as of 3 pairs are correctly match

=P(Ist correct , IInd correct and IIIrd correct)

=(1/3)×(1/2)×(1) = 1/6 = 0.16 (option-b).

(20)

Assuming that the children have not yet learned to associate the words with the images, the probability that they will match the three photos and three words correctly is same as above = 0.16 (option-b)

(21)

Now if X represents number of correct matches then E (X) is expected value of X

=0×P(X=0 correct) +1×P(X =1correct) + 2×P(X=2 correct) + 3×P(3=correct)

=0×(0.5) + 1×(0.33) + 2×(0.16) + 2×(0.16)

= 0 + 1×(1/3) + 2×(1/6) + 2×(1/6)

= 1/3 + 1/3 + 1/3

=3/3 = 1

Hence, E(X)=1 (option-b)

(22)

If X: number of correct matches then its variance is

V(X)=E(X²) - [E(X)]²

Since, E(X) = 3

And

E(X²)

= 0²×(0.5) + 1²×(0.33) + 2²×(0.1 6) +3²×(0.16)

=0 + 1+(1/3) - 4×(1/6) + 9×(1/6)

= 1/3 + 2/3 + 3/2

=4/3 + 3/2

=8/6 +9/6

=17/6

=2.833

So, E(X²) = 2.833

Now,

V(X) = E(X²) - [E(X)]² = 2.833 -(1)² = ​​​​​​1.833

Hence, V(X) = 1.833 (option-e)

(23)

If X: number of correct pairings then its standard deviation is

SD = √[V(X)] = √(1.833) = 1.354

Hence, standard deviation SD = 1.354 (option-e).

(24)

If we take a group of 30 children who have not yet learned to associate the words with the images then approximate children who would expected to correctly match a single word and photo pair is

=30×P(Only single pair is correctly match)

=30×0.33 = 30×1/3 = 10 (option-e)